Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.