tag:tagteam.harvard.edu,2005:/hub_feeds/3909/feed_itemsPeter Cameron's Blog2024-02-18T21:48:16-05:00TagTeam social RSS aggregratortag:tagteam.harvard.edu,2005:FeedItem/98165552024-02-18T21:48:16-05:002024-02-18T21:48:16-05:00Peter Cameron<p>
After recruiting Scott Harper to the team, we have finished the job of determining all the Cauchy numbers (these are the positive integers <i>n</i> for which there exists a finite list <b>F</b> of finite groups so that a finite group has order divisible by <i>n</i> if and only if some member of <b>F</b> is embeddable in it).
</p>
<p>
The answer is: <i>n</i> is a Cauchy number if and only if one of the following holds:</p>
<ul>
<li> <i>n</i> is a prime power; </li>
<li> <i>n</i> = 6; </li>
<li> <i>n</i> = 2<i>p<sup>a</sup></i>, where <i>p</i> is a Fermat prime greater than 3 and <i>a</i> is a positive integer. </li>
</ul>
<p>In the second and third cases, we can tell you the list <b>F</b>: for example, for <i>n</i> = 6, the list consists of the cyclic and dihedral groups of order 6 and the alternating group A<sub>4</sub>. (Of course, in the first case, the list consists of all groups of order <i>n</i>.)
</p>
<p>
The paper will appear on the arXiv on Tuesday, with the same number as before.</p>
Cauchy numbers: job doneAfter recruiting Scott Harper to the team, we have finished the job of determining all the Cauchy numbers (these are the positive integers n for which there exists a finite list F of finite groups so that a finite group … <a href="https://cameroncounts.wordpress.com/2024/02/18/cauchy-numbers-job-done/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/98117472024-02-16T06:18:12-05:002024-02-16T06:18:12-05:00Peter Cameron<p>
Anatoly Vershik died the day before yesterday.
</p>
<p>
As I have told here, he was the person who told me about the Urysohn space. I had given a talk at the ECM in Barcelona on the countable random graph, and after it he approached me and asked “Do you know about the Urysohn space?” We wrote a paper on it, extending some of my results on the random graph. Indeed, the Urysohn space has many different Abelian group structures.
</p>
<p>
He also made my two trips to St Petersburg possible, by having birthdays in 2004 and 2014 (actually his birthdays were in late December the previous year). I learned so much at these conferences, and had a small adventure when I came to leave after the first one.</p>
Anatoly VershikAnatoly Vershik died the day before yesterday. As I have told here, he was the person who told me about the Urysohn space. I had given a talk at the ECM in Barcelona on the countable random graph, and after … <a href="https://cameroncounts.wordpress.com/2024/02/16/anatoly-vershik/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/97985042024-02-11T10:59:47-05:002024-02-11T10:59:47-05:00Peter Cameron<p>
Richard Parker died last month. Now only two of the authors of the ATLAS of finite groups remain, the two Robs.
</p>
<p>
I knew Richard, but perhaps not well enough to write anything appropriate as a tribute. But I recommend you take a look at Rob Wilson’s <a href="https://robwilson1.wordpress.com/2024/01/27/richard-parker/">account</a> on his blog.</p>
Richard ParkerRichard Parker died last month. Now only two of the authors of the ATLAS of finite groups remain, the two Robs. I knew Richard, but perhaps not well enough to write anything appropriate as a tribute. But I recommend you … <a href="https://cameroncounts.wordpress.com/2024/02/10/richard-parker/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/97574992024-01-24T22:07:05-05:002024-01-24T22:07:05-05:00Peter Cameron<p>
Following on from the earlier post, the new version of the paper has just gone on the arXiv: <a href="https://arxiv.org/abs/2311.15652">2311.15652</a> (version 2).
</p>
<p>
If we say that <i>n</i> is a Cauchy number if there is a finite set <b>F</b> of finite groups, all with orders divisible by <i>n</i>, such that every group with order divisible by <i>n</i> must contain a group in <b>F</b> as a subgroup, then our result is as follows:
</p>
<p>
Let <i>n</i> be the product of two distinct primes <i>q</i> and <i>r</i>. Then <i>n</i> is a Cauchy number if and only if one of the primes is 2 and the other is a Fermat prime.
</p>
<p>
This means that, in all other cases, there are infinitely many groups which are minimal with respect to containing the cyclic groups of orders <i>q</i> and <i>r</i>. The arguments fall into two quite different cases. Apart from the pair {3,5}, there are infinitely many simple groups with this property. But for {3,5}, there is only one simple group, namely A<sub>5</sub>. So instead we use a theorem of Gaschütz which guarantees that we can build an infinite sequence of extensions, each a Frattini extension of the one before, starting with A<sub>5</sub>. (This means that each group after the first has a normal 2-subgroup contained in its Frattini subgroup such that the quotient is the preceding group.) These do the job.
</p>
<p>
I learned of this theorem from my co-authors. So collaboration, and putting a paper on the arXiv, have very positive results!</p>
More on Cauchy numbersFollowing on from the earlier post, the new version of the paper has just gone on the arXiv: 2311.15652 (version 2). If we say that n is a Cauchy number if there is a finite set F of finite groups, … <a href="https://cameroncounts.wordpress.com/2024/01/24/more-on-cauchy-numbers/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/97476542024-01-18T19:57:58-05:002024-01-18T19:57:58-05:00Peter Cameron<p>
Before you think I have gone totally crackers:
</p>
<p>
Cauchy’s theorem says that a finite group whose order is divisible by a prime number <i>p</i> contains a subgroup which is cyclic of order <i>p</i>.
</p>
<p>
My co-authors and I have proved some similar results, of which the one referred to in the title is the following:
</p>
<p>A finite grup whose order is divisible by 6 contains a subgroup which is either cyclic of order 6, dihedral of order 6, or isomorphic to the alternating group of degree 4 (with order 12).
</p>
<p>
When the more general theorem is proved and the paper written, I hope to elaborate on this. But my question for now is: have you seen this before?</p>
Cauchy’s theorem for the prime 6Before you think I have gone totally crackers: Cauchy’s theorem says that a finite group whose order is divisible by a prime number p contains a subgroup which is cyclic of order p. My co-authors and I have proved some … <a href="https://cameroncounts.wordpress.com/2024/01/18/cauchys-theorem-for-the-prime-6/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/97416872024-01-14T16:23:37-05:002024-01-14T16:23:37-05:00Peter Cameron<p>
I have been honoured by an invitation to give the inaugural Simon Norton lecture at the London Institute for Mathematical Sciences on 12 February. The webpage is <a href="https://lims.ac.uk/event/a-monstrous-talent/">here</a>.
</p>
<p>
Of course there are other people who knew Simon better than I did. Nevertheless, I think I have something to say. It was a paper of mine, with Jean-Marie Goethals and Jaap Seidel, which (as far as I know) introduced the term <em>Norton algebra</em> for a commutative but non-associative algebra of the general type which Bob Griess later used to construct the Monster. Our reference for this is to a personal communication from J. H. Conway. Nearly half a century later, I cannot remember exactly what Conway said to me.
</p>
<p>
Anyway, I will be very glad to see you there if you can come.</p>
Simon Norton lectureI have been honoured by an invitation to give the inaugural Simon Norton lecture at the London Institute for Mathematical Sciences on 12 February. The webpage is here. Of course there are other people who knew Simon better than I … <a href="https://cameroncounts.wordpress.com/2024/01/13/simon-norton-lecture/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/95938502024-01-02T00:38:16-05:002024-01-02T00:38:16-05:00Peter Cameron<p>
The youth could not help breaking a rule of courtesy towards this heavily burdened and yet, as he felt, noble man by asking: “But tell me, I beseech you, why do you carry on such wars on your star? Who is to blame for them? Are you yourself in part responsible?”
</p>
<p>
The King seemed angered at this audacity and for a time stared at the messenger. But he could not continue to meet with his dark gaze the bright and guileless eyes of the stranger.
</p>
<p>
“You are a child,” said the King, “and there are things you cannot understand. War is no one’s fault, it occurs of itself, like storm and lightning, and all of us who have to fight wars, we are not their originators, we are only their victims.”
</p>
<p>
Hermann Hesse, Strange News from Another Star</p>
New YearThe youth could not help breaking a rule of courtesy towards this heavily burdened and yet, as he felt, noble man by asking: “But tell me, I beseech you, why do you carry on such wars on your star? Who is to blame for them? Are you yourself in part responsible?” The King seemed angered … … <a href="https://cameroncounts.wordpress.com/2024/01/01/new-year/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/93246982023-12-19T23:39:00-05:002023-12-19T23:39:00-05:00Peter Cameron<p>
Is computer typesetting a kind of programming?
</p>
<p>
One of the pioneers, Donald Knuth, clearly thought so. In <em>The TeXBook</em>, he gives TeX code for computing and typesetting the first thirty primes; apart from anything else, this demonstrates that TeX has the capacity to act as an all-purpose program.
</p>
<p>
But there is a rather significant difference.
</p>
<p>
A programmer requires the program to deliver the right answer: exactly, for a discrete mathematician, and to with a specified approximation, for a continuous mathematician. Computer typesetting doesn’t deliver an answer as such. Knuth realises this when <em>The TeXBook</em> ends with the admonition “GO FORTH and create <em>masterpieces of the publishing art</em>“. Despite Keats, truth and beauty are not quite the same.
</p>
<p>
It is still true, though, that computer typesetting can pose problems similar to those faced by programmers. This summer, I was editing a book with four substantial chapters written by different teams of authors. When I put the chapters together and compiled the book, I discovered that the diagrams in one chapter were wrong. After a lot of searching, I discovered that this was caused by a conflict between two packages used by different authors. The standard remedy is to load the packages in the other order. But this brought LaTeX to a screaming halt without typesetting anything.
</p>
<p>
The two conflicting packages were <tt>curves</tt> and <tt>tikz</tt>. The authors who had loaded the second of these made extensive use of it, but for the first package only two commands had actually been used. The solution wasn’t going to be to delve inside the packages and make changes, since the LaTeX file had to go off to the publisher.
</p>
<p>
I used plain TeX for several years before being (more or less forcibly) converted to LaTeX. (If you have looked at my Combinatorics book for CUP, you will have noticed that it doesn’t look like most textbooks produced with LaTeX – that is the reason.) So I delved into my knowledge of TeX. I loaded <tt>curves</tt>, renamed the two required commands with the TeX <tt>\let</tt> command, and then loaded <tt>tikz</tt>, then held my breath and compiled the book: it worked!
</p>
<p>
So no pretence that this was the “correct” programming to use here, but it happened to work to produce the book: perhaps not a masterpiece of the publishing art, but at least a beautiful book. (It should appear next year.)
</p>
<p>
I now have to face a somewhat similar problem. A coauthor just sent me a copy of the paper, so that it is now my responsibility to make some edits. I was quite horrified, when I looked through it, to find the string lcm<i>p</i>. (I am showing you just an approximation in HTML.) I checked the input: lcm was correctly defined as a math operator, so the output should have looked something like lcm <i>p</i>. (A math operator is designed to leave a certain amount of space between itself and its operand; this looks better and is easier to read and understand. All this is based on a combination of centuries of typesetters’ experience and research in neuropsychology.)
</p>
<p>
I now have to track down the problem.
</p>
<p>
I also noticed that the preamble to the paper loads several dozen packages, many of which I have never heard of. I suspect that one of these is causing the trouble, and my next task will be to find which one. At the same time, I also suspect that most of these packages are not necessary. I know that some authors start with a LaTeX template which loads every package they have ever heard of, just in case it is needed. So I might do some judicious filtering …</p>
Programming and typesettingIs computer typesetting a kind of programming? One of the pioneers, Donald Knuth, clearly thought so. In The TeXBook, he gives TeX code for computing and typesetting the first thirty primes; apart from anything else, this demonstrates that TeX has … <a href="https://cameroncounts.wordpress.com/2023/12/19/programming-and-typesetting/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/91064572023-12-04T06:46:00-05:002023-12-04T06:46:00-05:00Peter Cameron<p>
<img src="https://cameroncounts.files.wordpress.com/2023/12/dominic.jpg?resize=219%2C219&w=640" width="25%" alt="Dominic">
</p>
<p>
Dominic Welsh died late last week.
</p>
<p>
Dominic and I were tutors at Merton College, Oxford for nearly eleven years. I was the pure maths tutor and he was the applied maths tutor. But there was no other Oxford college where the mathematical interests of the pure and applied maths tutors were closer. College lunches are an opportunity to talk to colleagues in other areas; Dominic and I broke this convention to some extent by discussing our research, though he was in no way an antisocial mathematician!
</p>
<p>
He had a very great impact on the subject. Much of this came from his provocative conjectures, but the main influence was through his students. Take a look at the <a href="https://www.mathgenealogy.org/id.php?id=53162">Mathematical Genealogy page</a>: his students include Adrian Bondy, Peter Donnelly, Graham Farr, Geoff Grimmett, Colin McDiarmid, Criel Merino, James Oxley, Ken Regan, and David Stirzaker.
</p>
<p>
Among other things, he was the second chair of the British Combinatorial Committee, my predecessor-but-one in this role.</p>
Dominic WelshDominic Welsh died late last week. Dominic and I were tutors at Merton College, Oxford for nearly eleven years. I was the pure maths tutor and he was the applied maths tutor. But there was no other Oxford college where … <a href="https://cameroncounts.wordpress.com/2023/12/04/dominic-welsh/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90670722023-11-28T06:07:40-05:002023-11-28T06:07:40-05:00Peter Cameron<p>
Here is the second of these discussions of elementary problems on finite groups with unexpected ramifications. This topic was suggested to me by Hamid Reza Dorbidi form Jiroft, Iran.
</p>
<p>
Let <b>F</b> be a finite set of finite groups. A finite group <i>G</i> is a cover of <b>F</b> if every group in <b>F</b> is isomorphic to a subgroup of <i>G</i>; it is minimal if no proper subgroup of <i>G</i> is a cover, and minimum if no group of smaller order than <i>G</i> is a cover. Covers certainly exist; for example, the direct product of the groups in <b>F</b> is a cover of <b>F</b>. But we would like more economical covers, that is, minimal or (especially) minimum covers. Note that any cover contains a minimal cover, but not necessarily a minimum one.
</p>
<p>
A paper on this topic just went on the arXiv, <a href="https://arxiv.org/abs/2311.15652">2311.15652</a>. I am not going to tell you about everything in that paper; I want to highlight some questions we cannot answer.
</p>
<p>
Perhaps the major question we can’t answer is this. It is clear that a set <b>F</b> can have only finitely many minimum covers. Which sets <b>F</b> have infinitely many minimal covers? Consider the set consisting of two cyclic groups of prime orders <i>q</i> and <i>r</i>. If (<i>q,r</i>) = (2,3), there are only three minimal covers, the cyclic group of order 6, S<sub>3</sub>, and A<sub>4</sub>. But if <i>q</i> and <i>r</i> are both greater than 5 there are infinitely many. So the question is quite subtle!
</p>
<p>
A set of nilpotent groups has a nilpotent minimum cover, and a set of <i>p</i>-groups has a minimum cover which is a <i>p</i>-group. But we do not know whether a set of abelian groups has a minimum cover which is abelian. We do, on the other hand, know the order of the smallest abelian group covering a set of abelian <i>p</i>-groups; I’ll describe this shortly.
</p>
<p>
Also, we do not know whether a set of <i>p</i>-groups can have infinitely many minimal covers, even if we strengthen the definition, and say that a cover is strongly minimal if it has no proper subgroup or quotient which is a cover.
</p>
<p>
We could also ask whether a set of soluble groups has a minimum cover which is soluble. But this question has a simple negative answer. The alternating group of degree 4 and the dihedral group of order 10 are both subgroups of A<sub>5</sub>, but no other group of equal or smaller size contains both of them.
</p>
<p>
If <b>F</b> is a set consisting of two non-abelian simple groups, then a minimum cover is either their direct product or simple. Both cases can occur:</p>
<ul>
<li> the minimum cover of PSL(2,8) and A<sub>5</sub> is their direct product; </li>
<li> the minimum cover of PSL(2,7) and A<sub>6</sub> is A<sub>7</sub>. </li>
</ul>
<p>This suggests a few questions:</p>
<ol>
<li> The pairs of simple groups of the same order are all known. Can any of these be minimum covers of a set of two simple groups? </li>
<li> Do there exist three simple groups with the property that the product of the orders of the first two is the order of the third? If so, can both of these groups of the same order be minimum covers of the set consisting of the first two?
</li>
<li> Is it the case that, for “most” pairs of simple groups, the unique minimum cover is their direct product? </li>
</ol>
<p>
Another type of question concerns minimum <i>n</i>-covers; these are minimum covers for the set of all groups of order <i>n</i>. Note that Cayley’s theorem shows that the symmetric group S<sub><i>n</i></sub> is an <i>n</i>-cover, but usually not the smallest such. For example, if <i>n</i> is a power of a prime <i>p</i> (and <i>n</i>>2), then the Sylow <i>p</i>-subgroup of S<sub><i>n</i></sub> is a smaller <i>n</i>-cover.
</p>
<p>
If <i>n</i> is a power of the prime <i>p</i>, then the order of a minimum <i>n</i>-cover is a power of <i>p</i>. But which power? In particular, is there an upper bound for the size of a minimum <i>p<sup>m</sup></i>-cover of the form <i>p</i><sup><i>F</i>(<i>m</i>)</sup>, where the function <i>F</i> is independent of <i>p</i>? If so, what is its order of magnitude? (We have a quadratic lower bound, but no upper bound since we do not even know whether such a function exists. The upper bound given by the Sylow <i>p</i>-subgroup of S<sub><i>n</i></sub> is not of this form: it is <i>p</i><sup>(<i>p<sup>m</sup></i>−1)/(<i>p</i>−1)</sup>. (This gives the right answer for 4, but for 8 it gives 128 whereas the correct value is 32.)
</p>
<p>
On the subject of small <i>p</i>-groups, we know that for odd primes <i>p</i>, the minimum size of a <i>p</i><sup>3</sup>-cover is either <i>p</i><sup>6</sup> or <i>p</i><sup>7</sup>, but do not know which it is. This question should perhaps be resolvable. I will sketch here the proof that there is no cover of order <i>p</i><sup>5</sup> for <i>p</i> > 3. Such a group would have nilpotency class at most 4, and so would be a regular <i>p</i>-group. In such a group, the elements of order 1 or <i>p</i> form a subgroup <i>A</i>. Since the group must contain both the elementary abelian group of order <i>p</i><sup>3</sup> and the non-abelian group of this order with exponent <i>p</i>, the order of <i>A</i> is at least <i>p</i><sup>4</sup>. But <i>G</i> also contains the cyclic group <i>C</i> of order <i>p</i><sup>3</sup>, and the intersection of <i>A</i> and <i>C</i> has order at most <i>p</i>, so their product has order at least <i>p</i><sup>6</sup>.
</p>
<p>
Computation shows that there are minimal 27-covers of order 729; indeed there are 16 such groups.
</p>
<p>
Note that, if <b>F</b> is a set of abelian <i>p</i>-groups, then the order of the (unique) smallest abelian group containing them is known. The algorithm to find the group is as follows. Write each group in <b>F</b> as a sum of cyclic <i>p</i>-groups of non-increasing order, say of orders <i>p</i><sup><i>a</i>(1)</sup>, …, <i>p</i><sup><i>a</i>(<i>k</i>)</sup>. We can assume that we have the same value of <i>k</i> for all the groups, by adding “dummy” copies of the trivial group. Let <i>b</i>(<i>j</i>) be the maximum of all the values <i>a</i>(<i>j</i>) for fixed <i>j</i>, over all the groups in <b>F</b>. Then the smallest abelian group containing all the groups in <b>F</b> is the sum of cyclic groups of orders <i>p</i><sup><i>b</i>(1)</sup>, …, <i>p</i><sup><i>b</i>(<i>k</i>)</sup>; and this is its canonical form, since the <i>b</i>s will be non-increasing.
</p>
<p>
Using this, we can write down the order of the unique smallest abelian group containing all abelian groups of order <i>p<sup>m</sup></i>: It is <i>p</i><sup>Φ(<i>m</i>)</sup>, where Φ(<i>m</i>) is the sum of the floor of <i>m</i>/<i>k</i> as <i>k</i> runs from 1 to <i>m</i>. The function Φ was studied by Dirichlet, and has many interpretations; its order of magnitude is roughly <i>m</i> log <i>m</i>, and its values appear as sequence A006218 in the <a href="https://oeis.org">On-Line Encyclopedia of Integer Sequences</a>, where many references can be found.
</p>
<p>
We do not know whether there is a smaller non-abelian <i>p</i>-group which embeds all abelian groups of order <i>p<sup>m</sup></i>.</p>
Covers of groupsHere is the second of these discussions of elementary problems on finite groups with unexpected ramifications. This topic was suggested to me by Hamid Reza Dorbidi form Jiroft, Iran. Let F be a finite set of finite groups. A finite … <a href="https://cameroncounts.wordpress.com/2023/11/28/covers-of-groups/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90653442023-11-27T12:20:44-05:002023-11-27T12:20:44-05:00Peter Cameron<p>
In his youth, Paul Simon thought of himself as a poet:</p>
<blockquote><p>
On a tour of one-night stands
My suitcase and guitar in hand
And every stop is neatly planned
For a poet and a one-man band
</p></blockquote>
<p>
He thought about his craft:</p>
<blockquote><p>
My thoughts are scattered and they’re cloudy
They have no borders, no boundaries
They echo and they swell
From Tolstoy to Tinkerbell
Down from Berkeley to Carmel
Got some poems in my pocket and a lot of time to kill
</p></blockquote>
<p>or, rather negatively given that poetry should communicate,</p>
<blockquote><p>
I have my books
And my poetry to protect me
I am shielded in my armor
</p></blockquote>
<p>
And surprisingly often he describes problems with the process:</p>
<blockquote><p>
And the song I was writing is left undone
I don’t know why I spend my time
Writing songs I can’t believe
With words that tear and strange rhymes
</p></blockquote>
<p>or</p>
<blockquote><p>
The poet writes his crooked rhyme
Holy holy is his sacrament
Thirty dollars pays your rent
On Bleecker Street
</p></blockquote>
<p>or</p>
<blockquote><p>
My life seems unreal, my crime an illusion
A scene badly written in which I must play
</p></blockquote>
<p>or again</p>
<blockquote><p>
My words like silent raindrops fell
And echoed in the wells of silence
</p></blockquote>
<p>
These songs included much of his best work at the time. But then, after a while, he was able to write</p>
<blockquote><p>
Funny how my memory slips
While looking over manuscripts
Of unpublished rhyme
Drinking my vodka and lime
</p></blockquote>
<p>And after that, this theme becomes much less prominent in his work.
</p>
<p>
For me, things were somewhat similar. Like many people, I wrote poetry in my youth. Julian Jaynes says something like “Poems are rafts grasped at by men drowning in inadequate minds”, but I think I knew from early on that one of the main reasons was to practise my writing, so that when I had something to say I could say it clearly. When Bob Dylan renounced the over-elaborate imagery of <em>Blonde on Blonde</em> for the clean simplicity of <em>John Wesley Harding</em>, I took that as a role model.
</p>
<p>
Could Simon’s experience happen in mathematics? It is possible to imagine that an important mathematical truth is expressed in “words that tear and strange rhymes”. More worryingly, an argument written in the most elegant style could be wrong, and we may be less likely to see the mistake because the writing is so good.
</p>
<p>
Recently I have been struggling with a proof in which the trivial case fought back and I had to devise an absurdly complicated argument to cope with it, but all the rest came out quite easily, with the inequalities falling into place. So there can be a mismatch between subject and style, which is not entirely our fault.</p>
“Words that tear and strange rhymes”In his youth, Paul Simon thought of himself as a poet: On a tour of one-night stands My suitcase and guitar in hand And every stop is neatly planned For a poet and a one-man band He thought about his … <a href="https://cameroncounts.wordpress.com/2023/11/27/words-that-tear-and-strange-rhymes/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90463272023-11-20T21:54:54-05:002023-11-20T21:54:54-05:00Peter Cameron<p>
This post, in a sense, follows on from the previous one, though I won’t explain the connection just yet.
</p>
<p>
Here is another property of permutation groups which occurred to me last night. I have no idea yet whether it is worth anything at all, but I am open to ideas.
</p>
<p>
Let <i>G</i> be a group with a normal subgroup <i>N</i>. We say that the extension <em>splits</em> if <i>N</i> is complemented in <i>G</i>, that is, there exists a subgroup <i>H</i> of <i>G</i> such that <i>NH</i> = <i>G</i> and <i>N</i>∩<i>H</i> = {1}. If this happens, then <i>G</i>/<i>N</i> is isomorphic to <i>H</i>.
</p>
<p>
So here is the new definition. Let <i>G</i> be a permutation group on a set Ω. I will say that <i>G</i> is <em>coherent</em> if, for any subset Δ of Ω, the setwise stabiliser of Δ splits over the pointwise stabiliser.
</p>
<p>
Not every permutation group is coherent. Let <i>G</i> be a group which does not split over a normal subgroup <i>N</i>, and let Ω be the disjoint union of the groups <i>G</i> and <i>G</i>/<i>N</i>, with the action of <i>G</i> by right multiplication. Let Δ be the coset space of <i>G</i>/<i>N</i>. Then the setwise stabiliser of Δ is <i>G</i> while the pointwise stabiliser is <i>N</i>. The smallest example is the cyclic group of order 4, acting with two orbits of sizes 4 and 2.
</p>
<p>
But some of our favourite examples are coherent. Examples include regular and Frobenius groups, the symmetric and alternating groups. (Coherence for the alternating groups takes a little thought; the others are clear.)
</p>
<p>So what would I like to do?
</p>
<p>
First and foremost, show that all groups in some class are coherent, or give a classification of those that are not. This might be possible for transitive groups, but is maybe more likely for primitive groups.
</p>
<p>
And, in parallel with this, connect the property of coherence to other notions in permutation groups or elsewhere in mathematics.
</p>
<p>
Suggestions welcome!</p>
Coherent permutation groupsThis post, in a sense, follows on from the previous one, though I won’t explain the connection just yet. Here is another property of permutation groups which occurred to me last night. I have no idea yet whether it is … <a href="https://cameroncounts.wordpress.com/2023/11/20/coherent-permutation-groups/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90327552023-11-15T12:51:14-05:002023-11-15T12:51:14-05:00Peter Cameron<p>
A paper with this title has just gone on the arXiv, <a href="https://arxiv.org/abs/2311.07995">2311.07995</a>.
</p>
<p>
The random graph, discussed at length here earlier, has the property of <em>homogeneity</em>: any isomorphism between finite subgraphs can be extended to an automorphism of the whole graph. Now unfortunately there are very few finite homogeneous graphs. They were all found by Tony Gardiner in the 1970s. The only ones are disjoint unions of complete graphs of the same size and their complements, the 5-cycle, and the line graph of <i>K</i><sub>3,3</sub> (the 3×3 grid).
</p>
<p>
In 1992, Udi Hrushovski established a remarkable result which goes a long way towards rectifying this. He showed that, for every finite graph <i>G</i>, there is a finite graph <i>H</i> containing <i>G</i> as an induced subgraph, such that every partial automorphism of <i>G</i> (isomorphism between induced subgraphs) extends to an automorphism of <i>H</i>. The property is called <em>Extension Property for Partial Automorphisms</em>, or EPPA for short, and the graph <i>H</i> is called an <em>EPPA witness</em> for <i>G</i>. (This would be trivial if there were enough finite homogeneous graphs that every finite graph is embeddable as induced subgraph in a homogeneous graph; but this is far from true!)
</p>
<p>
The obvious questions are: How large does an EPPA witness have to be? And what other classes of structures have EPPA witnesses?
</p>
<p>
On the second question, quite a bit is known: some kinds of things do, some don’t, and in somce cases (such as tournaments) the answer is unknown.
</p>
<p>
For the first question, we have a new graph-theoretic parameter to study, the EPPA number of a graph (the smallest number of vertices of an EPPA witness). The paper on the arXiv, by David Bradley-Williams, Jan Hubička, Matěj Konečný and me, has some results on this, including upper and lower bounds for the smallest EPPA witness for every <i>n</i>-vertex graph, roughly 2<sup><i>n</i></sup>. Also, we have a good estimate for the almost sure size of the smallest EPPA witness for an <i>n</i>-vertex random graph.
</p>
<p>
In fact David and I had been working intermittently on EPPA for several years. I want to give a bit of a teaser for our work, which might soon be forthcoming.
</p>
<p>
The main observation is that the smallest EPPA witness <i>H</i> for a graph <i>G</i> has a lot of symmetry. If we take the number of vertices to be as small as possible, then <i>H</i> is vertex-transitive, and is either vertex-primitive or a cover of a graph having vertex-primitive group with <i>G</i> as an induced subgraph. If we further require that <i>H</i> has the smallest number of edges, then it is arc-transitive. So we would expect to be able to tap into the large amount of knowledge about such graphs.
</p>
<p>Indeed we can. As an example, consider the <i>n</i>-cycle graph. It is not too hard to show that the line graph of <i>K<sub>n</sub></i>, with <i>n</i>(<i>n</i>−1)/2 vertices, is an EPPA witness. Using the Classification of Finite Simple Groups, we can show that it is the smallest EPPA witness for all but finitely many <i>n</i>. We do not know the complete list of exceptions, but it is non-empty: for example, the 4-cycle and 5-cycle are homogeneous and so are their own EPPA witnesses, and the smallest EPPA witness of the 6-cycle is the 3×3 grid.</p>
EPPA numbers of graphsA paper with this title has just gone on the arXiv, 2311.07995. The random graph, discussed at length here earlier, has the property of homogeneity: any isomorphism between finite subgraphs can be extended to an automorphism of the whole graph. … <a href="https://cameroncounts.wordpress.com/2023/11/15/eppa-numbers-of-graphs/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90291442023-11-14T00:59:40-05:002023-11-14T00:59:40-05:00Peter Cameron<p>
A small story about how these things work. This is a paper with six authors, which has just been accepted for publication in the diamond journal <em>Combinatorial Theory</em>.
</p>
<p>
James East and three other semigroupists, Des FitzGerald (my classmate at the University of Queensland some time ago), James Mitchell (my colleague now in St Andrews), and Thomas Quinn-Gregson, were working on the problem: every finite semigroup can be embedded in a complete transformation semigroup (the analogue of Cayley’s Theorem for groups); but what is the smallest degree of a complete transformation semigroup embeddiing a given semigroup <i>S</i>?
</p>
<p>
The question for groups has been studied by various people including, most successfully, Babai, Goodman and Pyber in the 1990s. Roughly paraphrased, the obstruction to a group having a fathful permutation representation of small degree is the existence of an abelian subgroup of small index.
</p>
<p>
Perhaps one cannot expect general results of this sort for semigroups; in any case, the authors decided to restrict their attention to specific semigroups. Most of the effort went into <em>variants</em> of complete transformation semigroups. (The variant of the semigroup <i>S</i> defined by the element <i>a</i> is the semigroup with the same underlying set but with multiplication given by <i>x</i>*<i>y</i> = <i>xay</i>.) For the full transformation semigroup <i>T<sub>n</sub></i>, the minimum degree was known to be between <i>n</i> and 2<i>n</i>−1; the paper considerably improves these bounds.
</p>
<p>
However, as a sideline, they tackled some other semigroups such as bands. In this case, the problem is entirely combinatorial, and James thought that I might be interested in it. I was able to translate it into a hypergraph colouring problem, and come up with a conjecture about the value, with some evidence for it. So I decided to pass it on, and publicised it on this blog. This had the intended result: Luke Pebody got interested in it, and proved the conjecture, making up the set of six authors.
</p>
<p>
PS: I learned the word “semigroupist” from João Araújo, a semigroupist who can use the English language in constructive ways.</p>
ConnectionsA small story about how these things work. This is a paper with six authors, which has just been accepted for publication in the diamond journal Combinatorial Theory. James East and three other semigroupists, Des FitzGerald (my classmate at the … <a href="https://cameroncounts.wordpress.com/2023/11/13/connections/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90093252023-11-07T03:52:16-05:002023-11-07T03:52:16-05:00Peter Cameron<p>
Fairly recently I submitted a referee report to one of the big academic publishers. At the end of the process, it asked me whether I wanted it to tell ORCiD that I had submitted a referee report. Instinctively I said no. Later, I started wondering why. It seemed with hindsight that I had good reasons.
</p>
<p>
As is commonplace now, we know that we do the real work for academic publishing: we do the research, we write the papers and submit them, we edit the journals, we referee the papers. The publishers simply publish them (often screwing them up in the process), collect the APCs, and pay very large dividends to their shareholders.
</p>
<p>
As I have also said more than once, I don’t need to do this now, and usually don’t do it for single-author papers: I put them on the arXiv and maybe publish them in a diamond journal. But often I have junior coauthors for whom publication in a paid-for journal matters, so I go through the charade.
</p>
<p>
For there is one thing that publishers give us for our money (although, of course, diamond journals do also): the ability to tick the box saying “refereed article”. This matters to the bean-counters who judge our research and decide on things like promotion. Without the referees, the publishers couldn’t offer that. So they need the referees.
</p>
<p>
So how do I justify my instinctive reaction to that question? Two reasons, that I can see:</p>
<ul>
<li> The refereeing process is supposed to be anonymous. I know from teaching cryptography that any small leakage of information can help to break the cipher. </li>
<li> The more important reason, I think, is that the publishers know they need us and have to try to keep us on board. Despite their huge profits, they can’t pay us a fair price for our time, so they have to find some artificial way to give us a pat on the back, and this seems to be the current way of doing that. </li>
</ul>
<p>
I am getting old, and I can’t work at the rate that I once could. I get far more referee requests than I can possibly deal with. A fair proportion of these are outside my area of expertise, and another fair proportion are not well written. Both of these categories take much longer to deal with. So I admit it, I am not as efficient a referee as I once was, and all too often miss deadlines.
</p>
<p>
But I keep doing it for several reasons. First, and most important, collegiality: my papers have to be refereed for my co-authors’ sake, even if not for mine, and this is just return for that (though I referee far more papers than I write). But as well, occasionally I can help the authors: maybe I know something they don’t, or maybe I can spot the trick that will solve one of their problems, and I am under the illusion that it helps them if I point this out. Sometimes, too, this leads to new collaborations and new friendships. And finally, many of the editors are my friends, and I know well the trouble they go through finding referees, so I try to help as I can.
</p>
<p>
Incidentally, I see that the term “reviewer” has almost completely supplanted “referee”. It is probably more appropriate. A referee, in a sporting event, makes decisions which can only be overruled with great difficulty, whereas an academic referee simply advises the editors. I have argued with referees in the past, and I am glad to say that usually editors have made sensible decisions in these cases. The referee’s word is not law, even though it must be treated with respect.</p>
RefereeingFairly recently I submitted a referee report to one of the big academic publishers. At the end of the process, it asked me whether I wanted it to tell ORCiD that I had submitted a referee report. Instinctively I said … <a href="https://cameroncounts.wordpress.com/2023/11/07/refereeing/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90029612023-11-05T01:38:11-05:002023-11-05T01:38:11-05:00Peter Cameron<p>
News came that Gary Seitz died at the end of October.
</p>
<p>
My one paper with Gary (also Cheryl Praeger and Jan Sazl) was the proof of Sims’ conecture, stating that the order of the point stabiliser in a primitive group is bounded by any non-trivial subdegree. The proof uses the Classification of Finite Simple Groups; I recall that some of us were a bit hesitant to claim that the conjecture was proved, but Gary said we should bite the bullet. It is a theorem, we should use it.
</p>
<p>
Let me share another story. One time I was visiting the University of Oregon to work with Bill Kantor. At the same time, Martin Liebeck was visiting to work with Gary. One evening we went out to dinner together. I was put in mind of two long-married couples: Bill and I were deep in our work, Gary and Martin deep in theirs.</p>
Gary SeitzNews came that Gary Seitz died at the end of October. My one paper with Gary (also Cheryl Praeger and Jan Sazl) was the proof of Sims’ conecture, stating that the order of the point stabiliser in a primitive group … <a href="https://cameroncounts.wordpress.com/2023/11/04/gary-seitz/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/90029622023-11-05T01:38:36-05:002023-11-05T01:38:36-05:00Peter Cameron<p>
Last Friday, former Australian prime minister Scott Morrison did a question and answer session in St Andrews, hosted by Stephen Gethins of the International Relations department (and former SNP MP for north-east Fife). Rosemary and I went along; we were by some margin the oldest people there. I expected to disagree with him on various things, but we turned out to be more in agreement than not.
</p>
<p>
I will try to represent his views accurately below, but my memory may fail sometimes.
</p>
<p>
Almost the entire session was Q&A, mostly questions from the floor but some sent in advance. The first one was I think from Stephen and asked him what he thought he had done well and what less well during his time as Prime Minister (2018 to 2022). He dodged the second part, but remarked that it had been an interesting time, with a number of difficult issues, notably Covid, which he thought Australia had dealt with well and for which he took some credit.
</p>
<p>
Here are some of his views on things as they came over from answers to questions.</p>
<ul>
<li> The age of globalisation is over; we now have powerful nations (he named China, Russia and Iran) trying to create hegemonies in their parts of the world. (He described Xi as the most Marxist leader of China since Mao.)</li>
<li> Related to that, what should Western democracies do to help small developing countries? He was clear that we should not try to make them just like us; we should offer them the opportunity to be whatever it is that they want to be. In particular, if they need energy to develop their industries, we should not try to enforce what source of energy they should use. (In describing Chinese policy, he used the word “debt”, but did not elaborate, although much could be said.)</li>
<li> Australia’s economy has always been based on trade, and so trading agreements are important to Australia. Contrast this with the USA, where there is a suspicion that trading pacts will cause American industries to be undercut by foreign competition. He cited the example of the Australian town of Newcastle, the first in the country to be based on heavy industry; now the industry has gone but the town has successfully re-invented itself. </li>
<li> He is opposed to “moral equivalence”, by which if a country is prepared to trade with us then we assume that the actions of its government are justified; he follows Ronald Reagan, who called out the Soviet Union. </li>
<li> On fuel, he is opposed to the widespread introduction of wind and solar power, thinking that we should instead concentrate our efforts on the fuel of the future, which he considers to be hydrogen. He also pointed out that it is shale gas which has weaned the USA off coal. In addition, the technology for making solar panels is very polluting. He never mentioned nuclear. </li>
<li> He was saddened by the fact that Australia had a referendum on changing the constitution to provide representation for indigenous people, considering it divisive, and was pleased when the proposal was rejected. He claims that every indigenous community is different, and there is no uniform strategy that fits all. </li>
<li> He remarked on the huge change on the way we get our news now. He described social media as like “pissing in your wetsuit”, warm and “local” but unhealthy. He much prefers a debate between people with different views. (He is a powerful personality, so I imagine he often gets the better of such debates.) </li>
</ul>
<p>
Since I get the last word, I will take the opportunity of taking him to task over his energy policy. First, I think that the situation is so urgent that we can’t just sit back and wait ten years for new sources of energy to come onstream. I would have thought that this would be obvious to an Australian, in view of the devastating floods and fires recently. And second, his solutions are not solutons. It takes a lot of energy to make hydrogen, so this just pushes the problem somewhere else. And shale gas is itself a fossil fuel with the added disadvantage that its recovery leaves the underground spaces full of highly dangerous chemicals which are not going to stay there indefinitely!</p>
Scott MorrisonLast Friday, former Australian prime minister Scott Morrison did a question and answer session in St Andrews, hosted by Stephen Gethins of the International Relations department (and former SNP MP for north-east Fife). Rosemary and I went along; we were … <a href="https://cameroncounts.wordpress.com/2023/11/04/scott-morrison/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/89884932023-10-31T00:47:51-04:002023-10-31T00:47:51-04:00Peter Cameron<p>
Recently a couple of people have asked me questions, or suggested research topics, related to what appear to be elementary properties of finite groups, but on examination, show unexpected complexities. Part of the reason why group theory is such an endlessly fascinating topic. This one comes from Hiranya Kishore Dey in Bangalore, India. I hope to discuss the other one soon.
</p>
<p>
The paper we wrote is on the arXiv, <a href="https://arxiv.org/abs/2310.06516">2310.06516</a>. So you can read it if you want. There is an elephant in the room, which I will talk about later. (I hope this is an acceptable metaphor for a paper with an Indian co-author.)
</p>
<p>
Let <i>G</i> be a finite group of order <i>n</i>. Its order sequence, os(<i>G</i>), is the <i>n</i>-tuple of orders of the elements of <i>G</i>, arranged in non-decreasing order (so starting with 1 for the identity). The order sequences are partially ordered, where one sequence dominates another if every term of the first is at least as big as the corresponding term of the second. (Strictly speaking this is a partial order on sequences, not on groups, since there are pairs of groups with the same order sequence. It is a partial preorder on groups, but I will sometimes speak as if it were a partial order.)
</p>
<p>
There are two natural compositions on order sequences. One is the pointwise product, and the other is the pointwise least common multiple. (That is, if the <i>i</i>th elements of <i>G</i> and <i>H</i> are <i>a</i> and <i>b</i> respectively, then the <i>i</i>th element of their composition is <i>ab</i> for the product, lcm(<i>a,b</i>) for the lcm. These two operations coincide if the orders of the two groups are coprime. If they are not coprime, then the lcm gives the order sequence of the direct product, while the product is not realised by any group.
</p>
<p>
Moreover, if <i>G</i> and <i>H</i> have coprime orders, then the order sequence of their product dominates the order sequence of any extension of <i>G</i> by <i>H</i>.
</p>
<p>
There are parts of the poset of order sequences which have nice properties. For example, the order sequences of abelian groups of order <i>p<sup>m</sup></i> (for <i>p</i> prime) form a lattice isomorphic to the lattice of partitions of the integer <i>m</i> under the natural partial order on partitions, a very well studied lattice.
</p>
<p>
The property of nilpotency is characterised by the order sequence (in the sense that if a group is nilpotent then so is any other group with the same order sequence). If <i>n</i> is such that there exists a non-nilpotent group of order <i>n</i>, then there is a non-nilpotent group whose order sequence is dominated by the order sequence of every nilpotent group.
</p>
<p>
There are connections with graphs as well (of course). If two groups have isomorphic power graphs, then their order sequences are equal; and if their order sequences are equal, then their labelled Gruenberg–Kegel graphs are equal.
</p>
<p>
Now to that elephant. When Hiranya first contacted me, he asked me if I could prove that the order sequence of the cyclic group dominates the order sequence of any other group of the same order. If this were true, then a number of results in the literature would be easy consequences, while others could be substantially improved. Although this attractive conjecture is true for groups with order up to 1000, and the weaker statement that the order sequence of the cyclic group is maximal in the domination order is easily proved, we have been unable to show the general version. Help welcome!</p>
Ordering groups by element ordersRecently a couple of people have asked me questions, or suggested research topics, related to what appear to be elementary properties of finite groups, but on examination, show unexpected complexities. Part of the reason why group theory is such an … <a href="https://cameroncounts.wordpress.com/2023/10/30/ordering-groups-by-element-orders/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/87504602023-10-06T15:06:19-04:002023-10-06T15:06:19-04:00Peter Cameron<p>
Following the combinatorial design challenges, here are three questions on Latin squares.
</p>
<p>
A <em>Latin square</em> is an <i>n</i>×<i>n</i> matrix with entries from an alphabet of size <i>n</i> (typically the integers from 1 to <i>n</i>) such that each letter appears once in each row and once in each column. A <em>transversal</em> is a set of <i>n</i> cells, one in each row, one in each column, and one containing each letter. An <em>orthogonal mate</em> of a Latin square is a partition of the cells into transversals. (If we assign a new letter from a different alphabet to each part, we obtain a second latin square with the property that any pair of letters, one from each alphabet, occur at a unique position.)
</p>
<p>
<b>Problem 1</b> (Ryser’s and Brualdi’s conjectures) Ryser conjectured that any Latin square of odd order has a transversal; Brualdi conjectured that any Latin square of order <i>n</i> has a partial transversal of size <i>n</i>−1. These are still open, though bounds on the defect of a partial transversal have recently been improved by Peter Keevash, Alexey Pokrovskiy, Benny Sudakov and Liana Yepremyan.
</p>
<p>
<b>Problem 2</b> Find the limiting proportions (assuming they exist) of Latin squares which have (a) a transversal and (b) an orthogonal mate. Find good asymptotics for the rate at which these limits are approached.
</p>
<p>
My guess for the limits would be 1 and 0 respectively.
</p>
<p>
<b>Problem 3</b>: a derangement problem. The permutation of columns mapping the first row of a Latin square to the second is a derangement (that is, has no fixed points): this is immediate from the definition. Given a derangement, we can take the first two rows of a potential Latin square to be the identity permutation and this derangement, and ask in how many ways it can be completed to a Latin square. My conjecture is that is surprisingly close to constant, independent of the chosen derangement. In other words, the ratio of the maximum to the minimum number of completions of a derangement tends very rapidly to 1 as <i>n</i> increases. A solution to the problem should include a good estimate for the rate of convergence!
</p>
<p>
For investigating this, note that conjugate derangements have the same number of completions. So we could ask whether the number of Latin squares having a derangement with given cycle structure is very nearly proportional to the number of derangements with this cycle structure. For <i>n</i> = 4, the double transpositions have twice as many completions as the 4-cycles; so although there are twice as many 4-cycles as double transpositions, the total numbers of squares are equal for the two types (288 each). Even for <i>n</i> = 7, the answer is very different.
</p>
<p>
I do not know the precise numbers. Latin squares of order 7 where determined in a paper by H. W. Norton in 1939; although this may not be reliable I thought it would give a good indication. But the paper was published in the <em>Annals of Eugenics</em>, and because Latin squares of order 7 are such a potentially racist and offensive topic, the publishers make it difficult to get hold of. When I did, I found that the pages of tables were rotated and it would have been a terrible job to extract the information I wanted.
</p>
<p>
So instead I turned to the Markov chain, devised by Jacobson and Matthews, for choosing a random Latin square. The limit is the uniform distribution on Latin squares, but unfortunately it is not known how long you should run the chain until it is close to the limit. So I did 100000 trials, with 100 steps of the chain between each sample. These results should maybe taken with a grain of salt…
</p>
<p>
There are four cycle structures of derangements: 7, 5.2, 4.3, and 3.2.2. The proportions of each, to four places of decimals, are 0.3883, 0.2718, 0.2265 and 0.1132. In my experiment the numbers I obtained were 38610, 27264, 22854 and 11272. Pretty close, and this is only three more than a value for which the ratio of max to min number of extensions is 2.
</p>
<p>
Nicholas Cavenagh, Catherine Greenhill and Ian Wanless have results on this which, though impressive, are some way from what I believe to be the case.</p>
Some challenges on Latin squaresFollowing the combinatorial design challenges, here are three questions on Latin squares. A Latin square is an n×n matrix with entries from an alphabet of size n (typically the integers from 1 to n) such that each letter appears once … <a href="https://cameroncounts.wordpress.com/2023/10/06/some-challenges-on-latin-squares/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/86896212023-10-01T14:38:53-04:002023-10-01T14:38:53-04:00Peter Cameron
<p>Since WordPress changed their editor and it is no longer possible to write posts in HTML, I have to find a new solution to posting mathematics.</p>
<p>What I have done this time is to put it on my web page on GitHub (<a href="https://cameroncounts.github.io/web/">https://cameroncounts.github.io/web/</a>) and link it from here.</p>
<p>On Thursday last week, Peter Keevash came to talk to the Pure Maths Colloquium in St Andrews about some of his really astonishing breakthroughs on the asymptotic existence and enumeration of combinatorial designs, including t-designs, Latin squares, Sudoku squares, and many others. But there are some problems he cannot solve, including Ryser’s and Brualdi’s conjectures on Latin squares. So I jotted down three problems as a challenge for magicians wielding the new powerful methods. The first, of course, is the existence of projective planes, which he mentioned in his talk. The second is a generalisation of the existence of resolvable designs; his methods may well solve this already. The third is a perhaps lesser known problem due to Mohammed Aljohani, John Bamberg and me, to which Steiner sytems are the proposed answer rather than the subject of the problem.</p>
Challenges in combinatorial design theorySince WordPress changed their editor and it is no longer possible to write posts in HTML, I have to find a new solution to posting mathematics. What I have done this time is to put it on my web page … <a href="https://cameroncounts.wordpress.com/2023/10/01/challenges-in-combinatorial-design-theory/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/85346702023-09-24T01:25:00-04:002023-09-24T01:25:00-04:00Peter Cameron
<p>This was just reported to me by a group theorist.</p>
<p>He asked ChatGPT whether Aut(G) wr Sym(n) is a subgroup of Aut(G^n). It was honest enough to say that it didn’t know (though it said it in a rather roundabout way).</p>
<p>Then he asked whether equality holds when G is a finite simple group. This time it was more definite; it said No. This seemed to be the result of confusion between the automorphism group and the outer automorphism group, together with some waffle masquerading as a proof.</p>
A cautionary taleThis was just reported to me by a group theorist. He asked ChatGPT whether Aut(G) wr Sym(n) is a subgroup of Aut(G^n). It was honest enough to say that it didn’t know (though it said it in a rather roundabout … <a href="https://cameroncounts.wordpress.com/2023/09/23/a-cautionary-tale/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/85028972023-09-12T20:46:14-04:002023-09-12T20:46:14-04:00Peter Cameron
<p>News came yesterday that Donald Keedwell has died.</p>
<p>Donald Keedwell was the author (with J. Dénes) of a classic, <em>Latin Squares and their Applications</em>, first published in 1974 with a foreword by Paul Erdős, and still the best source of reference. In my work with Rosemary Bailey, Cheryl Praeger and Csaba Schneider, we used a theorem of Frolov for which the best reference we could find was Dénes and Keedwell. The book is his lasting memorial.</p>
<p>Donald was a stalwart of British combinatorics. He organised the British Combinatorial Committee at the University of Surrey in 1991, and was a member of the committee for some time, holding the office of Secretary (if my memory is correct) for a while.</p>
Donald KeedwellNews came yesterday that Donald Keedwell has died. Donald Keedwell was the author (with J. Dénes) of a classic, Latin Squares and their Applications, first published in 1974 with a foreword by Paul Erdős, and still the best source of … <a href="https://cameroncounts.wordpress.com/2023/09/12/donald-keedwell/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/84981042023-09-11T02:48:33-04:002023-09-11T02:48:33-04:00Peter Cameron
<p>The order in which the authors of an academic paper are listed is a perpetual source of trouble, partly since the approved order various between disciplines, so the contribution of authors of interdisciplinary papers is likely to be misunderstood.</p>
<p>In some areas, authors are ordered by the relative importance of their contributions. This leads to comparing apples and chalk: how do you rate the person who had the idea against the student who did all the hard work carrying it out?</p>
<p>Sometimes the order has a coded significance known only to insiders, e.g. the last-named author is the director of the lab. (If you do this, take care: remember what happened to Elena Ceaucescu, director of the Romanian chemistry institute, and incidentally wife of the dictator of Romania.)</p>
<p>In mathematics we almost always use alphabetical order. But this is tough on people whose name begins with a letter late in the alphabet. My former colleague Rob Wilson lists the authors of the Online Atlas of Finite Group Representations in reverse alphabetical order, so that he can be first for a change.</p>
<p>Now, in a recent arXiv preprint (2304.01393), Erik and Martin Demaine discuss a solution to this problem: superimpose the authors’ names!</p>
<p>Read their paper: it gives a beautiful account of the philosophy and mechanism of their method, including LaTeX code to format a BibTeX entry in this way.</p>
<p>Do take a look.</p>
Order of authorsThe order in which the authors of an academic paper are listed is a perpetual source of trouble, partly since the approved order various between disciplines, so the contribution of authors of interdisciplinary papers is likely to be misunderstood. In … <a href="https://cameroncounts.wordpress.com/2023/09/10/order-of-authors/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/84403942023-09-05T18:42:42-04:002023-09-05T18:42:42-04:00Peter Cameron
<p>Welcome a new diamond open access journal, <em>Innovations in Graph Theory</em>, which has just been announced. Ross Kang asked me to spread the word, which I am very happy to do.</p>
<p>The journal is based at the Mersenne Centre in Paris, and so is a stablemate of <em>Algebraic Combinatorics</em>. The webpage is <a href="https://igt.centre-mersenne.org/">https://igt.centre-mersenne.org/</a> . The first issue is expected next year.</p>
<p>As usual, I encourage everyone (especially those like me to whom impact factors no longer matter) to submit your best papers to diamond journals. Gold open access puts thousands of pounds/dollars/euros into the pockets of large publishers; even if you are not paying this yourself, somebody is (either it is paid by your grant, or it comes with a national agreement with a publisher; in either case, the taxpayer pays, and no matter what your politics, this is probably not what you want.</p>
New diamond journalWelcome a new diamond open access journal, Innovations in Graph Theory, which has just been announced. Ross Kang asked me to spread the word, which I am very happy to do. The journal is based at the Mersenne Centre in … <a href="https://cameroncounts.wordpress.com/2023/09/05/new-diamond-journal/">Continue reading <span>→</span></a>tag:tagteam.harvard.edu,2005:FeedItem/83126012023-08-16T09:36:28-04:002023-08-16T09:36:28-04:00Peter Cameron
<p>More sad news: Ian Macdonald has died. (To avoid confusion, this is the Ian Macdonald who introduced the Macdonald polynomials, not the one who taught me group theory.)</p>
<p>My first exposure to the ideas of algebraic geometry came at Oxford, from the Mathematical Institute lecture notes called “Notes on Commutative Algebra” by Atiyah and Macdonald, which became the famous book.</p>
<p>Ian Macdonald had a careful and precise style, enlivened by a sense of humour. In the introduction to “Symmetric functions and Hall polynomials”, he explains the difference between two conventions for writing Young tableaux and similar things by saying, “Francophones should read this book upside down in a mirror.” (I remember puzzling over this comment, since I was not familiar with the French convention. The difference between the labelling of elements of a matrix (or words in a left-to-right language) and Cartesian coordinates is a 90 degree rotation and does not require a reflection. I learned about Young tableaux using the matrix convention. I think that a different convention would require more changes than just reading the book upside down in a mirror. For example, the Robinson-Schensted algorithm involves “bumping” an element out of a row, so that it falls down to the row below and tries to find a place there. How would you describe that with a different convention?)</p>
<p>I will tell two anecdotes about Ian Macdonald. The first one he told me himself, and I witnessed the second.</p>
<p>He directed a programme at the Isaac Newton Institute. He arrived and was installed in a big office. A little later, the technician came running round: there was a problem on the network, a computer had crashed. Ian’s response: “That thing was sitting there humming; I found it distracting, so I turned it off.”</p>
<p>He was a plenary speaker at the British Combinatorial Conference at the University of Surrey. At that time, the medium of choice for giving a big lecture was overhead projector slides; but Ian preferred to write on the board in his elegant handwriting. Below the big screen were fixed blackboards running the entire width of the room. He began at the top left and worked his way across. At the point where he had filled all the boards and came back to the top left, he was about to generalise the whole set-up, so only minor changes were needed to what was already there. Given that he presumably had not seen the set-up of boards before arriving at the conference, this achievement would have required careful preparation!</p>
<p>He has left us with some conjectures about characters of finite groups, on which group theorists are making good progress. If I understand correctly, the truth of these conjectures follows from the truth of considerably stronger (and more complicated) conjectures about simple groups, which can probably be proved using the classification of finite simple groups. Whether Macdonald would regard this as a proof from the book, I do not know.</p>
Ian MacdonaldMore sad news: Ian Macdonald has died. (To avoid confusion, this is the Ian Macdonald who introduced the Macdonald polynomials, not the one who taught me group theory.) My first exposure to the ideas of algebraic geometry came at Oxford, … <a href="https://cameroncounts.wordpress.com/2023/08/16/ian-macdonald/">Continue reading <span>→</span></a>