Suppose $G$ is a finite group. We will say, that it force solvability if any finite group $H$, such that $G$ is isomorphic to its maximal proper subgroup, is solvable. Does there exist some sort of classification of such groups?

On one hand, all such groups have to be solvable. On the other hand, there are several large classes of such groups known.

One of them is yielded by a theorem from “A condition for the solvability of a finite group” by W. E. Deskins:

All groups of nilpotency class 2 force solvability

The other comes from a theorem by Thomas Browning:

If a finite group is nilpotent and all its $2$-subgroups are normal, then it forces solvability.

His proof is here:

Let $G$ be minimal such that $G$ is not solvable and such that $G$ contains a maximal subgroup $M$ that is nilpotent and whose 2-subgroups are normal. If $M$ contains a nontrivial normal subgroup $N$ of $G$ then $G/N$ contradicts the minimality of $G$. Thus, $M$ does not contain nontrivial normal subgroups of $G$. In particular, $N_G(P)=M$ for all Sylow $p$-subgroups $P$ of $M$. Then $P$ is a Sylow $p$-subgroup of $N_G(P)$ so $P$ is a Sylow $p$-subgroup of $G$. This shows that $M$ is a Hall subgroup of $G$.

If $P$ is a Sylow $p$-subgroup of $M$ and if $Q$ is a nontrivial normal subgroup of $P$ then $N_G(Q)=M$ which has a normal $p$-complement. For $p=2$, Frobenius' normal $p$-complement theorem gives that $G$ has a normal $p$-complement. For $p\geq3$, Thompson's normal $p$-complement theorem or Glauberman's normal $p$-complement theorem gives that $G$ has a normal $p$-complement (since you only have to consider characteristic $p$-subgroups).

Thus, for each prime $p$ dividing the order of $M$, $G$ has a normal $p$-complement. Then $M$ has a normal complement $N$ in $G$. Since $M$ is solvable but $G$ is not solvable, $N$ is not solvable. In particular, $N$ does not admit a fixed-point-free automorphism of prime order. If $m\in Z(M)$ has prime order then $C_N(m)$ is nontrivial. Then $C_N(m)M$ is a subgroup of $G$ that properly contains $M$ so $C_N(m)M=G$ by the maximality of $M$. Comparing cardinalities shows that $C_N(m)=N$ so $m\in Z(G)$. Then $\langle m\rangle$ is a nontrivial normal subgroup of $G$ contained in $M$ which is a contradiction.

It is also known, that if $H$ and $K$ force solvability, then $H \times K$ also does.

However, I do not know, whether there is anything else here...

  • $\begingroup$ This question was also asked on Math.SE: link $\endgroup$ Commented Aug 6, 2019 at 13:27
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    $\begingroup$ @MikkoKorhonen this is not the same question; it is rather a follow-up to the linked MathSE question (asked in September 2018 by Thomas Browning). $\endgroup$
    – YCor
    Commented Aug 6, 2019 at 13:40
  • $\begingroup$ @YaniorWeg Why is it that if $H$ and $K$ force solvability then $H\times K$ does? $\endgroup$ Commented Dec 6, 2019 at 2:51

1 Answer 1


I make a remark in the opposite direction. Let $p \geq 17$ be a Fermat or Mersenne prime, so that $X = {\rm PSL}(2,p)$ has a dihedral Sylow $2$-subgroup $D$ which is maximal. Let $d >1$ be a power of $2$, and let $Q$ be a transitive $2$-subgroup of $S_{d}$. Let $A= {\rm Aut}(X)$ and let $T$ be a Sylow $2$-subgroup of $A$. Then $T$ is a maximal subgroup of $XT$.

Using this action of $Q$ on $d$ points, form the wreath product $XT \wr Q$. Then $T \wr Q$ is a Sylow $2$-subgroup of $(XT) \wr Q$, and is a maximal subgroup of $(XT) \wr Q$.


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