Definition: Let $G$ be a group, and let $H \leq G$ be a subgroup. We say that $H$ is big in $G$ if for every intermediate subgroup $H \leq L \leq G$ there exists some $x \in L$ such that $\langle H \cup \{x\} \rangle = L$.

Question: Is there a big solvable subgroup in every finite group?

Motivation: By Theorem A in M.Aschbacher and R.Guralnik, "Solvable generation of groups and Sylow subgroups of the lower central series", in every finite group $G$ there exists a pair of conjugate solvable subgroup $H_1,H_2 \leq G$ such that $\langle H_1 \cup H_2 \rangle = G$. So if $x \in G$ is such that $xH_1x^{-1} = H_2$ then $\langle H_1 \cup \{x\} \rangle = G$. In my question I am asking for a generalization of this conclusion.

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    $\begingroup$ related: mathoverflow.net/questions/208776/… ($A_4^n$ is "big" in $A_5^n$ for every $n$). You could provide a little more about what you currently know about the question; for instance whether you know the answer for all finite simple groups. $\endgroup$
    – YCor
    Jun 9, 2015 at 10:09
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    $\begingroup$ This definition might not be universally accepted. I have once learned that a proper subgroup $H$ of $G$ is big if it intersects every conjugacy class. For instance the triangular subgroup is big in ${\bf GL}_n(\mathbb C)$. In a finite group, there does not exist such a big subgroup. $\endgroup$ Jun 9, 2015 at 11:24
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    $\begingroup$ No I think it's fine to ask for all finite groups, and mention that you don't know for finite simple groups. For instance for $\mathrm{Alt}_n$, you know for which values of $n$? $\endgroup$
    – YCor
    Jun 9, 2015 at 12:19
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    $\begingroup$ I don't think $A_6$, $A_7$ and $A_8$ are out of reach... $\endgroup$
    – YCor
    Jun 9, 2015 at 17:06
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    $\begingroup$ $A_6$, $A_7$ and $A_8$ all have maximal soluble subgroups. For example, $(S_4\times S_4)\cap A_8$ is a maximal intransitive subgroup of $A_8$. You can play a similar trick in $A_9$ by taking a maximal imprimitive group. $A_{10}$ is the smallest alternating group that doesn't have a maximal soluble subgroup. $\endgroup$
    – verret
    Jun 10, 2015 at 6:47

2 Answers 2


As a partial answer generalizing Pablo's example, the claim is true for a group with a $(B,N)$-pair for which $B$ is solvable. In this case, take $H:=B$. This recovers Pablo's example as $A_5$ acts doubly transitively on a set with 5 points, hence has a $(B,N)$-pair, with a solvable point stabilizer $B=A_4$. Other examples include e.g. general linear groups $GL_n(K)$.

See Wikipedia for the definition of a $(B,N)$-pair and the notation used in the sequel.

Indeed, any subgroup $L$ satisfying $B\leq L \leq G$ is a standard parabolic subgroup, $L=P_X$ for some subset $X$ of the index set of the Dynkin diagram of $G$. Write $X$ as the union of its connected components, $X=\cup C_i$, and for each $i$ choose some $g_i$ in the big cell ${B_X}^+{B_X}^-$. Then $g:=g_1\cdots g_n$ is as required, i.e. $B_g:=\langle B,g\rangle = L$. To see this, note that $B_g$ again is a standard parabolic subgroup and by the choice of $g_i$, which isn't contained in a proper parabolic subgroup of $P_{C_i}$, $B_g \cap P_{C_i} = P_{C_i}$.

  • $\begingroup$ Does this mean that you can somehow treat the case of $G = \mathrm{SL}_n(\mathbb{F}_q)$ where $q$ is a power of a prime > 5? $\endgroup$
    – Pablo
    Jun 10, 2015 at 22:27
  • $\begingroup$ Why is $g$ not in a proper parabolic? Does it come from, say, looking at to what Bruhat cell it belongs? $\endgroup$
    – LSpice
    Feb 19, 2021 at 16:28

Michio Suzuki proved that every finite group is generated by a pair of conjugate solvable subgroups. See http://projecteuclid.org/download/pdf_1/euclid.hokmj/1381517825 (Open access). The subgroup $S$ which Suzuki exhibits is not a priori "big" in your sense (as far as I can see), but may suggest an approach to the question.

Later edit: Ah, I see that the existence part of Suzuki's result is no stronger than that of Aschbacher and Guralnick, it's just that his subgroup $S$ has more properties which may be useful from an inductive point of view.


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