# Group extensions isomorphic as groups

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?

• Do you know an example of two different cohomology classes which are equivalent in this way? – Mark Grant Mar 31 '14 at 14:22
• Perhaps the simplest case is $H^2(Z_p,Z_p)$, for all nontrivial elements the associated group if $Z_{p^2}$. A most radical example is given in the answer of mathoverflow.net/questions/35649/… – César Galindo Mar 31 '14 at 14:49
• You should precise that $OpExt(G,A)$ is the group of classes of extension of $G$ by $A$ such that the induced action of $G$ onto $A$ is the one you started with. Ot – Joël Mar 31 '14 at 17:30

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 $(G,M)$.
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 $M$.