TITLE: Group extensions isomorphic as groups
QUESTION [5 upvotes]: Let $G$ be a group and $A$ a $G$-module. It well know that there is a group isomorphism between the second cohomologoy group $H^2(G,A)$ and the abelian group $OpExt(G,A)$ of classes of extension $$1\to A\to E\to G\to 1$$ of $G$ by the abelian group $A$. 
Define over $H^2(G,A)$ the equivalence relation where two cohomology classes are equivalent if  for the associated extensions $1\to A\to E_1\to G\to 1$, $1\to A\to E_2\to G\to 1$, $E_1\cong E_2$ as groups. 

Question: There is a systematic way in order to decide if two elements in $H^2(G,A)$ are equivalent?

REPLY [4 votes]: Define a homomorphism from the group-module pair $(G,M)$ to the group module pair $(H,N)$ to be a pair $(\phi,\psi)$, where $\phi:H \to G$ is a group homomorphism, $\psi:M \to N$ is a morphism of abelian groups, and $h\psi(m) = \phi(h)m$ for all $m \in M$, $h \in H$.  Then we get an induced homomorphism $H^k(G,M) \to H^k(H,N)$. (I hope I got that right!)
Now two extensions of $M$ by $G$ are isomorphic as groups (with an isomorphism that maps $M$ to $M$) if and only if the corresponding cohomology classes in $H^2(G,M)$ are equivalent under an automorphism of the group-module pair $(G,M)$.
This is something that can be computed, and has been used in practice, for example in the construction of the small groups library.
Note that this theory would not cover the possibility of an a group isomorphism from one extension ot $M$ by $G$ to another using an isomorphism that did not fix $M$.