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This is a question from a problem set on group cohomology, a subject I've just begun to learn.

Let $B$ be a finite group and $A$ be abelian. I am looking for two groups $G_1$ and $G_2$ such that $G_1$ and $G_2$ are isomorphic as groups but $$1\rightarrow A\rightarrow G_1\rightarrow B\rightarrow 1$$ and $$1\rightarrow A\rightarrow G_2\rightarrow B\rightarrow 1$$ are not isomorphic as extensions.

It has been suggested that I use $A=C_3^2$ and $B=C_2$. However, since the orders of $A$ and $B$ are relatively prime in this case, doesn't the Shur-Zassenhaus Lemma guarantee that the sequence splits so that there is only one extension? If this is the case, then how could we produce two non-isomorphic extensions? If someone could point out where I'm confused, I'd be very grateful.


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What do you mean exactly by "isomorphic extensions"? – DonAntonio Oct 23 '12 at 2:43
@DonAntonio: The extensions are isomorphic or equivalent if the diagram commutes. QiaochuYuan: Ah, I see. So, even if the sequence splits, the semidirect products are not unique because of the different actions one could use. – Alexander Sibelius Oct 23 '12 at 3:05
@Alexander: disregard my previous comment. You want to realize the same group as a semidirect product of $A$ and $B$ but in two different ways. – Qiaochu Yuan Oct 23 '12 at 3:06
Out of curiosity, might I ask a question? If the two extensions differ by an automorphism on $G_1$, can we call them different? If so, then of course any extension with non-trivial $\operatorname{Aut}(G_1)$ should do. But, if not, I think of this question as very strange, for we are asked to find two different "representations" of the same group$\ldots$ am I missing something here? – awllower Dec 8 '12 at 15:01
@AlexanderSibelius - You may want to look at the accepted answer to the related question. – chizhek Sep 30 '14 at 10:25

1 Answer 1

I think the "smallest" counter-example is the following :

Here I denote $\mathbb{Z}_k$ the group $\frac{\mathbb{Z}}{k\mathbb{Z}}$. Take $A:=\mathbb{Z}_2$, $G_1=G_2=G=\mathbb{Z}_4\times\mathbb{Z}_2$ and $B:=\mathbb{Z}_2\times \mathbb{Z}_2$ as abelian groups. You have two injections :

$$\alpha: A\rightarrow G $$

$$a\mapsto (0,a) $$

and :

$$\alpha': A\rightarrow G $$

$$a\mapsto (2a,0) $$

Those maps are clearly monomorphisms of groups. Furthermore it is not hard to see that $G/\alpha(A)=G/\alpha'(A)=B$. Hence we indeed get two extensions of $B$ by $A$ :

$$0 \rightarrow A \overset{\alpha}{\rightarrow} G \overset{\beta}{\rightarrow} B \rightarrow 0\text{ and }0 \rightarrow A \overset{\alpha'}{\rightarrow} G \overset{\beta'}{\rightarrow} B \rightarrow 0$$

Clearly $G_1$ and $G_2$ are isomorphic as abelian groups but the extensions cannot be isomorphic, if they were isomorphic then there would be an automorphism $\Phi$ of $G$ such that :

$$\Phi\circ \alpha =\alpha'$$

In particular :


But this impossible because if we define $a:=\Phi^{-1}(1,0)$ then :

$$\Phi(a+a)=(1,0)+(1,0)=(2,0) $$

Hence $a+a=(0,1)$ but if $a=(a_1,a_2)$ then $a+a=(2a_1,0)\neq (0,1)$. Hence such a $\Phi$ cannot exist, hence the extensions are not equivalent.

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