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Is it true that if $A$ and $B$ are two non-abelian finite simple groups, then the only finite group $G$ which has one copy of $A$ and one copy of $B$ as composition factors is $G = A \times B$? If not, could someone give a counterexample or even better, a reason why this isn't true?

Certainly, these are the two types of extensions we have to consider:

$$1 \to A \to G \to B \to 1$$ $$1 \to B \to G \to A \to 1$$

It seems true so for all the examples I tried, but I don't have definitive proof.

Shaun
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3 Answers3

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The answer to the question is yes, the only group with two nonabelian finite simple groups $A$ and $B$ as composition factors is the direct product $A \times B$.

The well-known Schreier Conjecture says that the outer automorphism group of any finite nonabelian simple group is solvable. The conjecture was finally confirmed by the Classification of Finite Simple Groups (of course it would have been much nicer if there a direct proof had been found).

So, in any semidirect product $A \rtimes_\phi B$, for a homomorphism $\phi:B \to {\rm Aut}(A)$, we must have ${\rm Im}(\phi) \le {\rm Inn A}$, and hence $A \rtimes_\phi B \cong A \times B$.

The smallest finite group in which all composition factors are nonabelian and which is not a direct product of simple groups is the wreath product $A_5 \wr A_5 \cong A_5^5 \rtimes A_5$of order $60^6$.

Derek Holt
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  • Thanks! If $\mathrm{Out}(A)$ is solvable how does it imply that $\mathrm{Inn}(\phi) \leq \mathrm{Inn}(A)$? (Sorry, I don't remember much about what solvability implies) –  May 11 '20 at 08:23
  • @Triskelion It's not $Inn(\varphi)$ but $Im(\varphi)$. As to why that's true, I suggest you try a little bit, and let us know where you get stuck. – verret May 11 '20 at 08:27
  • @Derek Holt You seem to assume that we have a semi-direct product. It would require only a little more work to deal with the case of a general extension A.B, which seems to be what the OP asked about. – verret May 11 '20 at 08:28
  • @verret Yeah, sorry typo. Anyway, I looked up the definition of solvable groups and they say that it is one whose derived series terminates in the trivial subgroup. So if $\mathrm{Out}(A)$ is solvable, and its derived series terminates in the trivial subgroup, what does it imply? Could you give some hints? –  May 11 '20 at 08:36
  • In the finite case, soluble case is equivalent to having all composition factors abelian. Such a group can't have a nonabelian simple group as a subgroup. For the rest of the hints, see my comment to the other answer. – verret May 11 '20 at 08:46
  • @verret Yes, sorry, I was in a hurry, and assumed that the extension was pslit. Thanks for completing the proof! – Derek Holt May 11 '20 at 11:32
  • One would expect that something similar would be true if instead both $A$ and $B$ are semisimple (i.e. direct products of non-abelian simple groups). By similar, I mean that at least the extension is split. That is not true, however. – the_fox May 11 '20 at 12:13
  • @the_fox The smallest nonsplit extension that I can think of has the structure $A_6^{30}.A_5$. (It exists because of the group $M_{10}$, which is a noncplit extension $A_6.2$ of order $720$.) – Derek Holt May 11 '20 at 12:43
  • @verret Could you tell me if $\mathrm{Im}(\varphi)$ is necessarily a simple group though? –  May 11 '20 at 16:10
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    Since $B$ is simple, either ${\rm Im}(\phi) =1$ or ${\rm Im}(\phi) \cong B$. In the first case we have the direct product and in the second case ${\rm Im}(\phi)$ is simple. – Derek Holt May 11 '20 at 16:14
  • @DerekHolt Thanks! –  May 11 '20 at 16:16
  • @DerekHolt I am not sure if that's the smallest. It's possible though. One interesting bit is that any non-abelian simple group can work as the top group, with the bottom being a direct product of copies of $A_6$. A crucial property is that every (non-abelian) simple group has a $C_2 \times C_2$ subgroup. This is proved in a paper by Gross and Kovacs (Theorem 4.7 here). – the_fox May 11 '20 at 16:41
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Here's an argument that doesn't assume that the extension is split. WLOG, let $A$ be a normal subgroup of $G$, with quotient isomorphic to $B$, both $A$ and $B$ being nonabelian simple (and everything finite).

Let $C$ be the centraliser of $A$ in $G$. Since $A$ is normal in $G$, so is $C$. Note also that $A\cap C=1$ hence $AC=A \times C$. It also follows that $C$ is isomorphic to a normal subgroup of $B$, so either $C=1$ or $C\cong B$. In the latter case, we get $G\cong A\times B$, as required. Otherwise, $C=1$ and by the N/C Theorem, $G$ embeds in $\mathrm{Aut}(A)$. In fact, a variation of this argument yields that $G/A$ embeds in $\mathrm{Out}(A)$. Now, as Derek pointed out, this $\mathrm{Out}(A)$ is soluble (by the Schreier Conjecture), which is a contradiction.

verret
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  • What is the NC theorem btw? –  May 11 '20 at 10:23
  • @Triskelion Try to make a little effort before you ask a question. This one is answered by googling "NC Theorem" and trying the first hit: https://en.wikipedia.org/wiki/Centralizer_and_normalizer – verret May 11 '20 at 23:25
  • If Out(A) is solvable, why can't G/A embed in it? –  May 12 '20 at 12:48
  • I think I understand the last argument now. 1. A simple solvable group is necessarily cyclic of prime order. 2. $G/A \cong B$ but $B$ is non-abelian and by assumption simple. 3. However, as subgroups of solvable groups ($\mathrm{Out}(A)$ is the solvable group in this context) are solvable, $B$ must be solvable as well. So contradiction! –  May 12 '20 at 13:44
  • The "variation of this argument" that you refer to is explained in this answer. –  May 12 '20 at 13:49
  • Why is $A\cap C = 1$ though? –  May 12 '20 at 13:54
  • Nevermind, I found the answer for why $A\cap C = 1$ here. –  May 12 '20 at 14:41
  • How exactly does it follow that $C$ is isomorphic to a normal subgroup of $B$? –  May 12 '20 at 16:24
  • Okay, found the answer here! –  May 12 '20 at 17:06
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As @Jim says in his comment, $A_5 \rtimes A_5$ works. To spell this out, we have $\operatorname{Aut}(A_5) \cong S_5$, so there is a (unique) nontrivial homomorphism $\varphi : A_5 \to \operatorname{Aut}(A_5)$. The semidirect product $A_5 \rtimes_\varphi A_5$ therefore fits in a short exact sequence $1 \to A_5 \to A_5 \rtimes_\varphi A_5 \to A_5 \to 1$, but $A_5 \rtimes_\varphi A_5 \not\cong A_5 \times A_5$.

Edit: to clarify, the uniqueness of $\varphi$ is not necessary for the argument, but it is cool that this is the unique smallest counterexample.

Correction This is wrong in two ways: first of all, the uniqueness of $\varphi$ is only up to an automorphism of $A_5$ (the image is unique). More importantly, $A_5 \rtimes_\varphi A_5 \cong A_5 \times A_5$, verified by GAP. I'm running a computer search for another counterexample, but this answer should be ignored for now.

diracdeltafunk
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  • Thats the easy part, the hard part is proving that $A_5 \rtimes A_5 \neq A_5 \times A_5$. – Jim May 11 '20 at 04:21
  • Well, only a trivial homomorphism $H \to \mathrm{Aut}(N)$ degenerates into the direct product, isn't it? That I think is a standard theorem. (Oh, but the resulting semidirect product might be isomorphic to the direct product I guess if it's an inner automorphism) –  May 11 '20 at 04:25
  • I think so, but I couldn't find a reference to double check. It's clear that the trivial hom gives you the direct product but the other direction is not so obvious. Especially since $A$ is not necessarily the only normal subgroup of $A \rtimes B$ that is isomorphic to $A$. – Jim May 11 '20 at 04:29
  • @Triskelion Only the trivial homomorphism $H \to \operatorname{Aut}(N)$ makes $H \rtimes G$ equal to the direct product, but it's conceivable that $H \rtimes G$ is isomorphic to $H \times G$ some other way (see https://math.stackexchange.com/q/201710/19006). I'm not sure if there's a theorem about this in general, and I thought I had a reason to believe $A_5 \rtimes_\varphi \not\cong A_5 \times A_5$, but I was wrong! I just checked by computer and they are indeed isomorphic. I'll retract my answer. – diracdeltafunk May 11 '20 at 05:12
  • @diracdeltafunk Thanks but please don't delete your answer though. Even wrong approaches are sometimes helpful! Which computer program are you using btw? –  May 11 '20 at 05:19
  • @Triskelion I'm using GAP – diracdeltafunk May 11 '20 at 05:23
  • @Jim Now I'm thinking the right question to ask might be whether any homomorphism $\varphi: H \to \mathrm{Aut}(N)$ is such that $\mathrm{Im}(\varphi) \leq \mathrm{Inn}(N)$? (When $H$ and $N$ are non-abelian simple groups.) –  May 11 '20 at 05:57
  • @Triskelion This is not the "right" question, it's just a different question. Anyway, before you ask it, I encourage you to try it on your own. Here's a hint: suppose $G$ is a group with a normal subgroup $K$ such that $G/K$ is soluble, and $H$ is a nonabelian simple subgroup of $G$, then $H\leq K$. (Consider the canonical projection.) Now try to see how this applies to this situation. – verret May 11 '20 at 08:43
  • @verret Oh, so $H = \mathrm{Im}(\phi)$, $K = \mathrm{Inn}(A)$ and $G=\mathrm{Aut}(A)$ in this context! Could you point me to a proof of the theorem in your hint though? –  May 11 '20 at 08:55
  • @Triskelion No, I think you should try it as an exercise first. I already gave you a few hints. – verret May 11 '20 at 08:57
  • @verret You mean your hint was an exercise? If we take it as a theorem, I can see how $\mathrm{Im}(\varphi) \leq \mathrm{Inn}(N)$ follows. –  May 11 '20 at 08:57
  • Yes, but you should be able to prove what I wrote as well, by, as I said, considering the canonical projection and the fact that soluble groups can't have nonabelian simple subgroups (which is another exercise I suppose). – verret May 11 '20 at 09:05
  • @verret Okay, I'm trying but I guess I'll need some help (I don't know much about the concept of solvability). By canonical projection, I suppose you mean something like a homomorphism $\pi: G \to G/K$. $K$ certainly lies in $\mathrm{Ker}(\pi)$ (but $\mathrm{Ker}(\pi)$ might be strictly larger than $K$ as well I think). But then how does $G/K$ being solvable and $H$ being simple non-abelian imply that $H$ lies within $K$ as well? –  May 11 '20 at 09:18