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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.

Thanks.

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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 at 10:25

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