For a group $G$ and subgroup $H$, consider the relation on $G$ defined $x \sim y$ if $H^x \cap H^y = 1$. This defines a graph on $G$.

It is always fairly symmetric: $N_G(H)$ acts on the left and $G$ on the right as graph automorphisms.

For some choices of $H\leq G$ the automorphism group of the graph is much (much) larger.

For $H$ a Sylow $2$-subgroup of $G=S_5$, the resulting graph has 3840 undirected edges on 120 vertices and is extremely symmetric: the automorphism group $X$ of the graph has structure $2.A_6.2.2^4.2.2^4.2.2^4.A_8^{15}$ and acts vertex, edge, and arc-transitively. It has two orbits of non-edges. It is highly connected, each pair of vertices has at least 32 common neighbors (and in the complement graph, each pair of vertices has at least 22 common neighbors). Neither it nor its complement is bipartite.

While working on graphs defined by slightly more complicated (and irrelevant) means, I was surprised to find that nearly all of the graphs were disjoint unions of complete graphs. This is the first example that is not in some sense “fully symmetric”, but I am a bit overwhelmed by how to study it.

How does $A_8^{15}$ act on $S_5$?

$|G|=8\cdot 15$ and $|H|=8$ but I'm not sure.

Does $X$ have a normal subgroup isomorphic $S_8^{15}$?

This would account for the middle 2s, but then what is up with them grouping into singles and quadruples, but the $A_8$ grouping into a 15-tuple?

What is the socle of the quotient of $X$ by $A_8^{15}$?

Is that $S_6$ acting on 3 copies of $2^4$? Are they all the same $2^4$ or are some conjugate under the outer automorphism of S6?

  • $\begingroup$ Short version: the automorphism group is a wreath product $S_8 \wr S_6(15)$. The $S_8$ part is arbitrary permutations of $H$. The $S_6$ acting on 15 points part is not yet clear, but true. $\endgroup$ Jul 15, 2013 at 20:25
  • $\begingroup$ Sylow-$2$ subgroup of $S_5$ is also Sylow-$2$ subgroup of $S_4$, so for $(G,H)=(S_4,Syl_2(S_4)$, what is the graph (only 24 vertices; so may be understood quickly). I want to get information about this graph (just for my understanding). $\endgroup$
    – Beginner
    Jul 16, 2013 at 8:20
  • $\begingroup$ In the $S_4$ case, the $G$-core of $H$ is non-trivial, so no intersections are trivial, so we get the null graph with 24 vertices and 0 edges. Its automorphism group is $S_{24}$. The fusion system of $G$ (partially) acting on $H$ is the same for $G=S_4$ and $G=S_5$, but the graphs are very different. In particular, the property of being an intersection of Sylow subgroups cannot be detected from the fusion. $\endgroup$ Jul 16, 2013 at 13:34

2 Answers 2


Any permutation of any of the cosets $Hx$ is an automorphism of the graph. So you can immediately see the subgroup $N = S_8^{15}$. I haven't attempted to prove it, but it seems clear that these cosets must form a system of imprimitivity for $X$, in which case the subgroup $N$ is indeed normal, so $X = S_{15} \wr P$, where $P$ is a transitive permutation group of degree 15 and, since it has $A_6$ as composition factor and order 720, must be $S_6$. (You can verify that from the transitive groups database.)

  • $\begingroup$ Thanks! Do you know why the $S_8/A_8$ would be grouped into those chief factors of size $2^4$ and $2$? I assume it has to do with the action of $S_6$ on 15 points, but I'm not sure how. $\endgroup$ Jul 15, 2013 at 20:06
  • $\begingroup$ Ok, I think I see it. The same grouping happens for $C_2 \wr S_6(15)$, and one gets different behavior for $C_p \wr S_6(15)$, so this is just decomposing a permutation representation into its irreducible modular representations. $\endgroup$ Jul 15, 2013 at 20:22
  • $\begingroup$ Yes you are just looking at the permutation module of $S_6$ in degree 15 over the field of order 2. You can study that using the MeatAxe, for example. It has three constituents of dimension 1, and three of dimension 3, two of which are isomorphic as modules. $\endgroup$
    – Derek Holt
    Jul 15, 2013 at 21:11
  • $\begingroup$ Thanks! I've verified by computer the group is $S_8 \wr S_6(15)$, so now I'm trying to prove things by hand so that I can see how they generalize. How did you know it was the full wreath product instead of just a subgroup? $\endgroup$ Jul 15, 2013 at 22:04

In progress:

Let $$X(G,H) = \left\{ f \in \newcommand{\Sym}{\operatorname{Sym}}\Sym(G) : H^x \cap H^y \neq 1 \iff H^{f(x)} \cap H^{f(y)} \neq 1 \right\}$$ be the group of graph automorphisms of the graph defined in the question.

Note that $\Sym\left({N_G(H)x}\right) \leq X(G,H)$ for every $x \in G$. This is simply because if $f \in \Sym\left({N_G(H)x}\right)$, then $H^{f(x)} = H^{nx} = H^x$.

If $T$ is a right transversal of $N_G(H)$ in $G$, then we get that the subgroup generated by the various $\Sym\left({N_G(H)x}\right)$ for $x \in T$ is a direct product and a subgroup of $X(G,H)$.

In general, it need not be a normal subgroup (take $G$ to be $S_4$ and $H$ to be a Sylow 2-subgroup, then $X(G,H)=\Sym(G)$ and $\Sym(8) \times \Sym(8) \times \Sym(8) \leq \Sym(24)$, but it is not normal).

If the embedding of $H$ in $G$ is “sufficiently rich” so that one can determine if $H^x = H$ based on the adjacencies of $x$, then $N_G(H)$ will generate a block system, and we will get a subgroup of a wreath product.

Now what permutes those blocks?

Why would we get the full wreath product?


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