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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?

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  • $\begingroup$ Do you know an example of two different cohomology classes which are equivalent in this way? $\endgroup$ – Mark Grant Mar 31 '14 at 14:22
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    $\begingroup$ 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/… $\endgroup$ – César Galindo Mar 31 '14 at 14:49
  • $\begingroup$ 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 $\endgroup$ – Joël Mar 31 '14 at 17:30
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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 $M$.

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  • $\begingroup$ The examples in the answer of mathoverflow.net/questions/35649/… show that your answer if not right. $\endgroup$ – César Galindo Mar 31 '14 at 18:12
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    $\begingroup$ Note that he explicitly states in his last paragraph that this approach does not cover automorphisms which don't fix the chosen normal divisor. The example you referred to is exactly that. $\endgroup$ – Achim Krause Mar 31 '14 at 18:33

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