The second week in Perth is over. Yesterday we went to Rottnest Island, named by the Dutch sea captain Vlaminck (“rats’ nest”) who mistook the quokkas for large rats; it is Wadjemup in the local Noongar language, and Rotto informally. It is an extraordinary place for birds. The picture shows a black-winged stilt (with what might be a plover of some sort behind), a pelican, baby oystercatchers, and a crested tern. And of course there were more quokkas than you could shake a stick at:
This one is a bit unusual; they are mostly very tame, especially round the settlement, but the first two we saw ran into the bushes to hide from us.
On Friday I gave a seminar on “Idempotent generation and road closures”; fortuitously, an email came round warning of further road closures in St Andrews that very day, and I was able to include it in my slides.
Without going into the connection with road closures, the question is: for which transitive permutation groups G is it true that, if O is an orbit of G on 2-subsets of the domain of G, and B a maximal block of imprimitivity for G acting on O, then the graph with edge set O\B is connected?
We know that such a group must be primitive, and must be basic (that is, preserves no Cartesian power structure on its domain). Moreover, João Araújo and I conjecture that, if G is a basic primitive permutation group such that G has no imprimitive subgroup of index 2 and also is not one of a class of examples related to the phenomenon of triality, then G has our “road closure” property.
Since I have been here, Cheryl Praeger and I have thought about one case of this problem, and have been thinking more generally about removing edges from a vertex-transitive graph to disconnect it.
Classical results of Mader and others assert that, if Γ is a vertex-transitive graph of valency k, then the edge-connectivity of Γ is k; this means that removing fewer than k edges do not disconnect the graph. Clearly it can be disconnected by removing the k edges through a vertex, to isolate that vertex. But the argument gives a little bit more. A set of k edges whose removal disconnects the graph must be either the edges through a vertex, or the k edges each with one end in a complete subgraph of size k. (Graphs with this property can be obtained from arc-transitive graphs by replacing each vertex with a clique of size k, each of whose vertices is on one of the edges through the original vertex.)
The second kind of disconnecting set cannot arise in an edge-transitive graph except for a cycle (with k=2) or a complete graph of size k+1, since in general the edges in the clique cannot be equivalent to the other edges under an automorphism.
The proof of this uses the concept of an atom. The boundary of a set of vertices is the set of edges having one end in that set; an atom is a set A of vertices whose boundary is minimal, and subject to that the size of A is minimal.
Cheryl went on to look at “pseudo-atoms”, sets of vertices whose boundary is minimal subject to being greater than k, and subject to this the size of the set is minimal. I will just say that if we add in the condition of edge-transitivity then such a set must be a single edge, with boundary of size 2k−2.
This already gives some information on the road-closure problem, and I hope we will have more progress to report soon. But Cheryl is off for a week in London and Paris on mathematics business …
These pages have not been maintained for a long time now, and I am sure that many of the links no longer work. But maintaining the pages is not just a case of bringing the links up to date (and deleting those which don’t work), but more importantly would require me to scour the web for new links to add.
I have always thought that a carefully curated page may be more useful than the result of a Google search, but the margin is becoming smaller. So reluctantly I have decided to remove these pages.
They will go at the end of this year. Please let me know if you have any comments or suggestions before then.
However, it hasn’t made the problems go away. See
https://pubpeer.com/publications/90BA3090D680D06326E43EC08CDEE8
for a depressing story.
Will this kind of thing push us to a situation where incomplete papers are posted on the arXiv just so as to claim priority? That would not be in anyone’s interest. But how to deal with the question otherwise?
Let us say that two elements of a finite group G are m-equivalent if one can be substituted for the other in any generating set for G without destroying the generating property. The “m” here stands for maximal subgroups, since two elements are m-equivalent if and only if they lie in the same maximal subgroups of G. In particular, the m-equivalence class of the identity is the Frattini subgroup, the set of non-generators. (This way of viewing m-equivalence leads to the best algorithm we know for computing it.)
The number of m-equivalence classes is an interesting invariant of the group G. We computed it for the first few symmetric groups, and this is now sequence A270534 in the On-Line Encyclopedia of Integer Sequences. The asymptotic behaviour of this sequence is an unknown problem, on which we make some comments in the paper.
But m-equivalence can be refined. Let us say that two elements are m-equivalent at level i if one can be substituted for the other in any i-element generating set. These equivalence relations, indexed by i, get finer as i increases. What can we say about their behaviour?
The number of level-i equivalence classes is 1 if i < d(G), the smallest number of generators for G, and is strictly greater than 1 if i = d(G). It is easy to see that, for i = d(G), the identity and all the elements of the generating set are pairwise inequivalent, so the number of equivalence classes jumps from 1 to at least d(G)+1. It might be interesting to find a good lower bound.
Another significant point is the value of i where the equivalence relation reaches its final value; we call this number ψ(G). Now one of the most interesting and surprising questions we have come up with, and not been able to answer, is:
Is it true that ψ(G) ≤ d(G)+1?
We have no counterexamples to this, and it follows from a result of Burness, Liebeck and Shalev that ψ(G) ≤ d(G)+5. If the answer to the above question is “yes”, then the number of equivalance classes takes at most three distinct values for any given group. This would also yield a remarkable dichotomy for finite groups, into those with ψ(G) = d(G), and those with ψ(G) = d(G)+1. Is there a more obvious property distinguishing the two types?
There are also connections with the generating graph. If d(G) ≤ 2, we can form a graph with vertex set G, where two vertices are joined if these two elements generate the group. Our research began from the observation that the automorphism group of the generating graph is huge (for the alternating group A_{5}, it has order 23482733690880). The reason is that any permutation which preserves the level-2 m-equivalence classes is an automorphism of the graph: two vertices in the same class have the same neighbours in the graph. In the paper we define a reduction process to a weighted graph whose automorphism group (at least in the case of almost simple groups) appears to be close to the automorphism group of G; the automorphism group of the full generating graph is the semi-direct product of a direct product of symmetric groups (on the level-2 m-equivalence classes) by the automorphism group of the weighted reduced graph, and the latter is equal to the automorphism group of G in many cases (though by no means all).
The vertex weight here is simply the cardinality of the equivalence class which is shrunk to give that vertex. An interesting question is to find examples where the reduced graph has automorphisms not preserving the weighting: we have a couple of examples, but the phenomenon is rare.
So there are several good problems to tackle here. Would you like to join in?
If you don’t need to read any more but simply to sign a petition, here is the link:
http://speakout.web.ucu.org.uk/no-cuts-no-confidence-at-university-of-leicester/
A few points occur to me.
So here is that petition again:
http://speakout.web.ucu.org.uk/no-cuts-no-confidence-at-university-of-leicester/
Yesterday we visited Yanchep National Park, north of Perth. I went there on my first visit to Perth in 1966 (when I played golf for the first and last time in my life – I was not very good at it!), and so I was looking forward to the chance to go back.
We were provided with maps by a friend, and had figured out that we could get there by public transport: bus into Perth, train to Butler, then another bus, which would leave us by the side of a busy road a few kilometres from the park. It turned out that there was no footpath beside the road, so we simply had to take our chances. A park employee later told us that the local public transport authority has no interest in providing transport to “tourist attractions”, and despite having been asked often to extend the bus route the short distance into the park, has refused to do so. How stupid is that?
Anyway, we walked by the side of the road, and finally reached a point where a path (part of several marked trails) passed adjacent to the road. So with a sigh of relief we took the path and had a wonderful walk. At the end, a park official berated us for not obeying the rules, which require you to fill in and sign a form before you set out on a trail (just in case the park catches fire!) Of course we hadn’t seen the rules. They are not on the map; if you come by car you will probably see them, but it seems people are not supposed to arrive at a national park on their own feet!
The trail we followed was the Ghost House Walk Trail, a circuit of about 12.4km. All that is left of the Ghost House is a doorway, but the path itself went through extremely varied terrain with a great variety of plant life including many lovely wildflowers. Since I posted some Kings Park wildflowers last week, I thought I would concentrate on animal (including bird and insect) life this week. As well as these, we saw many other birds including ringneck parrots, cockatoos, galahs, crows, willy wagtails, and Pacific black ducks.
As a final nod to Nicolas Baudin, the cockatoos we saw were probably Baudin’s black-cockatoo, though they may have been Carnaby’s.
See Yanchep while you can. A block of land right next to the park has been bought by a developer; the whole area is being bulldozed and turned into Perth suburbs, and no doubt it won’t be long before Yanchep is reduced to an isolated urban green space. (Of course it may be easier to reach by public transport then!)
This week, I gave a colloquium talk on “The random graph and its friends“, and Rosemary a seminar talk on “Circular designs balanced for neighbours at distances one and two“.
In addition, we got started on some research. I will just highlight one of the stories here.
In our first meeting, Michael reported that, after he had talked about derangements in Iran, he had been asked to what extent the derangements in a transitive group determine the group.
Jordan proved in 1872 that a finite transitive permutation group of degree n > 1 necessarily contains a derangement (an element with no fixed points). In fact it contains many derangements: Arjeh Cohen and I proved in 1992 that at least a fraction 1/n of the elements of the group are derangements. But I vaguely remembered seeing, at some time in the distant past, a theorem that said:
Theorem
Let G be a transitive finite permutation group of degree greater than 1, and H the subgroup generated by the derangements in G. Then:
Now my memory is not failing too badly: I was able to find a reference, Lemma 6.3 in a paper by H. Zantema, “Integer valued polynomials over a number field”, Manuscripta Math. 40 (1982), 155–203. But, before I did that, I was able to reconstruct the proof, and to add a bit more: with the same hypotheses,
In this form we see that Frobenius groups form a special case. In a Frobenius group G, every non-identity element fixes 0 or 1 point; the theorem of Frobenius states that the derangements together with the identity form a normal subgroup H. In this case, H_{1} is the trivial group, and its orbits are just the points; by definition, G_{1} permutes them semiregularly. So Frobenius groups give examples with H a proper subgroup of G.
Here is the proof of the theorem. Let π(g) be the number of fixed points of the permutation g. Suppose that the subgroup H generated by the derangements has d orbits.
By the Orbit-Counting Lemma,
∑ {π(g)−1 : g∈G} = 0,
∑ {π(g)−1 : g∈H} = (d−1)|H|.
So
∑ {π(g)−1 : g∈G\H} = −(d−1)|H|.
In this equation, the left-hand side is at least zero (since the only permutations giving negative contributions are the derangements, which are all in H), whereas the right-hand side is at most zero. So both must be zero, from which we conclude that d = 1 and also that all elements with π(g) ≠ 1 are in H.
For the last part, we use a slight variant. Let r_{G} and r_{H} be the permutation ranks of G and H, the numbers of orbits on ordered pairs. Then
∑ {(π(g)−1)^{2} : g∈G} = (r_{G}−1)|G|,
∑ {(π(g)−1)^{2} : g∈H} = (r_{H}−1)|H|.
Since all elements of G for which the summand is non-zero are in H, the two sums are equal. We conclude that
(r_{G}−1)|G : H| = r_{H}−1,
from which the second part of the theorem follows easily.
There are other examples apart from Frobenius groups. The unique almost-simple doubly transitive group, PΓL(2,8) (degree 28) has the property that its derangements generate PGL(2,8), a subgroup with index 3; the three orbitals are permuted transitively by the field automorphism.
We expect to have substantially more to report on this problem soon!
My St Andrews problems are here, and are in pretty good shape (there are only 14 of them so this wasn’t a very big job).
However, the many problems I collected at Queen Mary will need more work. The index page to these problems is here.
If you have any information about these problems having been solved, even if you think I already know it, please let me know. Email me at St Andrews with any information you might have. Credit will be given!
Added later And here is a summary of my conjectures (also a bit in need of updating!)
So yesterday, of course, we went walking in Kings Park. One of our leaflets describes Kings Park as the largest city park in the world; Wikipedia is more modest, calling it “one of the largest inned-city parks”. It is less than half the size of Richmond Park in London, and just over half the size of the Adelaide parklands. But size isn’t everything; Kings Park is a wonderful place.
Western Australia is famous for its wildflowers, at their best in the spring; Kings Park has a good selection of these:
We move on to the next stage of the trip, from Adelaide to Perth.
This was only my second visit to Adelaide. I intended to have a holiday, in the hope of seeing off the last vestiges of shingles. I did quite well: walking in the Adelaide Hills (Black Hill, Mt Lofty, and Waite where I saw the rainbow lorikeets above), the Park Lands all round the city, the River Torrens Linear Park, and the beach from the mouth of the Torrens down to Glenelg; the State Library, Art Gallery, Zoo, Botanic Gardens, and Central Markets; Port Adelaide, with the Maritime and Railway Museums and a walk to Semaphore and Outer Harbor; and a weekend in Robe with stops in various places including two spots on the Coorong. Some eating and drinking of course. Even some mathematics, which I hope to write about soon, though there are still a few bugs to be ironed out.
But I am not sure about throwing off the shingles. Things seem to move much more slowly. I can spend a morning writing a couple of references now, and this wears me out so that I get little done in the afternoon. Things are improving but so slowly.
So if you are waiting for me to do something, bear with me a little longer.
Anyway, we are now in Perth, and the real work is about to begin …