# condition for short exact sequence of groups to be isomorphic

Let $G$ be a group and $K_1,K_2$ be two distinct normal subgroups of $G$. We have two short exact sequences:

$$1 \to K_1 {\rightarrow} G {\rightarrow} G/K_1 \to 1$$

$$1 \to K_2 {\rightarrow} G {\rightarrow} G/K_2 \to 1$$

Assuming that we have an isomorphism $f: G/K_1 \stackrel{\cong}\to G/K_2$, under what additional conditions the two short exact sequences are isomorphic ?

I think, if the two sequences correspond to central extension then we have conditions based on the cohomology group $H^2(G,K_1)$ and $H^2(G,K_1)$. I do not assume however that we have central extensions.

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What is your definition of an isomorphism of exact sequences? – Julian Kuelshammer Jan 21 '13 at 20:20
For me the two short exact sequence are isomorphic if there are isomorphisms of group $g:K_1 \to K_2$ and $h: G\to G$ such that if we join the two exact sequences I write above by $f,g,h$, then the resulting diagram commute. (Sorry I was not able to draw the diagram so I write in letter). – user59152 Jan 21 '13 at 21:48
According to the 5-Lemma it would suffices to have an isomorphism $K_1\to K_2$ and a morphism $G\to G$ such that the diagram commutes, but I guess you want some weaker condition. – Julian Kuelshammer Jan 21 '13 at 21:53
Your accounts have been merged. To avoid creating duplicate accounts in the future, please register your account. – Qiaochu Yuan Jan 22 '13 at 5:39
A necessary and sufficient condition is for there to be an isomorphism $K_1 \to K_2$ that extends to an automorphism of $G$. I don't think there is much more you can say in general. So you need to provide more information on the problem you are trying to solve. – Derek Holt Jan 22 '13 at 9:10