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It's a long time I am not working with groups. Maybe you know the answer to these questions.

Suppose that $1\rightarrow A\xrightarrow{i_1} B_1\xrightarrow{\pi_1}C\rightarrow 1$ and $1\rightarrow A\xrightarrow{i_2} B_2\xrightarrow{\pi_2} C\rightarrow 1$ are two short exact sequences of (not necessarily abelian) groups. Suppose we know that they are isomorphic.

a) Is there an isomorphism $\beta:B_1\rightarrow B_2$ such that $(id_A,\beta,id_C)$ is an isomorphism between the two exact sequences? ( I mean, such that $\beta i_1=i_2$ and $\pi_1=\pi_2\beta$).

b) If no, then what about the case where the sequences split?

Thanks in advance.

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  • $\begingroup$ You mean $B_1$ and $B_2$ are isomorphic? As the sentence stands, “they” seems referring to the short exact sequences, rather than to their middle terms. $\endgroup$
    – egreg
    Commented Oct 16, 2013 at 17:17
  • $\begingroup$ I thought "they" was intended to apply to the sequences, meaning there are isomorphisms between the three components giving a commutative diagram. $\endgroup$
    – Derek Holt
    Commented Oct 16, 2013 at 21:38

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If I am understanding your question correctly, then the answer is no even for split extensions. For example, $A$ could be the additive group of a vector space on which the action of $C$ makes $A$ into a $C$-module. There could be an outer automorphism of $C$ that does not preserve the module. In that situation, if your given isomorphism between the exact sequences involved such an outer automorphism of $C$, then there would be no isomorphism $B_1 \to B_2$ that induced the identity on $C$.

As a specific example, take $C={\rm PSL}_3(2)$ with $N$ elementary abelian of order $8$, and $C$ inducing the natural module action, and use the duality outer automorphism of $C$. This does not preserve the module but maps it to its dual.

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  • $\begingroup$ Let $B=\mathbb{Z}_2^3\rtimes PSL_3(2)$ and let $\gamma:PSL_3(2)\rightarrow PSL_3(2)$ be the duality outer automorphism (could you give the definition of this automorphism?). So, you are saying that there is not an isomorphism $\beta:PSL_3(2)\rightarrow PSL_3(2)$ such that $\gamma\pi=\pi\beta$. Am I right? (here $\pi:B\rightarrow C$ is the projection). $\endgroup$
    – Diego
    Commented Oct 16, 2013 at 18:48
  • $\begingroup$ You mean $\beta:B \to B$? Yes, that's right! The isomorphism $\gamma$ is induced by the map $A \mapsto {\rm Transpose}(A^{-1})$ for matrices $A \in {\rm GL}_3(2)$. (In fact ${\rm GL}_3(2) \cong {\rm PSL}_3(2)$.) $\endgroup$
    – Derek Holt
    Commented Oct 16, 2013 at 21:35

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