The maximal subgroups of $A_{7}$ are $A_{6}$ (index 7), $S_{5}$ (index 21), $(A_{4} \times A_{3}):2$ (index 35), and $GL_{3}(2)$ (index 15).

If $G$ is a nonabelian finite simple group and all maximal subgroups of $G$ have odd index, does it follow that $G \cong A_{7}$?

I don't know the Lie type groups well enough to answer this for myself, but here I'll prove it for the alternating groups (and a case-by-case check works for the sporadic groups):

Assume that $n \geq 5$. If $n+1$ is not a power of 2, then choose an $r < \frac{n}{2}$ such that $\binom{n}{r}$ is even; then the stabilizer of an $r$-element subset of $\{ 1, \ldots, n \}$ in $A_{n}$ is a maximal subgroup of even index. Otherwise, write $n = 2^{k}-1$. Since we are assuming $n \neq 7$, we say that $k \geq 4$. Then $GL_{k}(2)$, in its action on vectors of $\mathbb{F}_{2}^{k} \setminus \{ 0 \}$, is a subgroup of $A_{n}$. Since $GL_{k}(2)$ is a proper subgroup of $A_{n}$, consider a maximal subgroup $M$ containing $GL_{k}(2)$.

Claim. $[A_{n}:M]$ is even.
Proof of Claim. $GL_{k}(2)$ acts doubly transitively on the indices, so $M$ does too. Therefore $M$ acts primitively on the indices. If $[A_{n}:M]$ were odd, then every 2-element of $A_{n}$ would be conjugate to an element of $M$; in particular, $M$ would contain permutations of cycle shape $(2,2)$. But $n = 2^{k}-1 \geq 15$ and, on more than $8$ points, the only primitive permutation groups containing a permutation of cycle shape $(2,2)$ are the symmetric and alternating groups. This is a contradiction because we assumed $M < A_{n}$.


1 Answer 1


There is an answer to the more general question of which finite (almost) simple groups have any maximal subgroup of odd index in the paper

M. Liebeck and J. Saxl, The Primitive Permutation Groups of Odd Degree J. London Math. Soc. (1985) s2-31 (2): 250-264.

They give a complete list of all groups of Lie type with this property. There are not so many examples of groups of Lie type that have some subgroup of odd index, and and there are no examples of groups of Lie type in which all maximal subgroups have odd index.

They refer to a different paper for the sporadic groups, but you could check them in the ATLAS.

  • 4
    $\begingroup$ It's worth mentioning that Kantor proved the same result independently. The paper is called Primitive permutation groups of odd degree, and an application to finite projective planes... $\endgroup$
    – Nick Gill
    Sep 30, 2016 at 11:20
  • 4
    $\begingroup$ ... Note, too, that both Kantor and Liebeck&Saxl missed one example: the group ${^2G_2}(q)$ contains a maximal subgroup of odd index with structure $(2^2\times D_{\frac12(q+1)}):3$. This subgroup is given in Kleidman's paper which classifies the maximal subgroups of ${^2G_2(q)}$. $\endgroup$
    – Nick Gill
    Sep 30, 2016 at 11:21
  • $\begingroup$ BTW, your answer misses the example given in the question -- you should write "in no cases, except $A_7$, do all maximal subgroups have odd index". $\endgroup$
    – Nick Gill
    Sep 30, 2016 at 11:58
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    $\begingroup$ But that statement is referring specifically to groups of Lie type. I am sure the statement is true also for sporadic groups, but I didn't check. I will clarify. $\endgroup$
    – Derek Holt
    Sep 30, 2016 at 12:00
  • 1
    $\begingroup$ I think some word is missing in "examples of groups of Lie type that have some subgroup of index" $\endgroup$
    – YCor
    Sep 30, 2016 at 20:40

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