# Under what conditions does the second cohomology preserve injectivity?

Suppose G, A, and B are abelian groups with $i:A \to B$ an injective homomomorphism. Consider the groups $H^2(G,A)$ and $H^2(G,B)$ for the trivial action of G on A and B. i induces an injective homomorphism:

$$i_*: H^2(G,A) \to H^2(G,B)$$

The map $i_{\ast}$ is not always injective. For instance, setting A as the integers embedded in B the rationals, we see that the target group of $i_*$ is always trivial but the source group is often nontrivial (for instance, when G is finite cyclic).

Question: Under what conditions is $i_*$ injective and under what conditions is it an isomorphism? More specifically I am interested in the case where:

G is a finite abelian p-group and A and B are both finite cyclic p-groups with the order of A dividing the order of B. Here, p is a prime number.

Generic note: Note that the induced map at the level of cocycles (before going down to cohomology) is injective, but the trouble arises because it seems possible that some cocycles that were not coboundaries become coboundaries on enlarging the group of coefficients.

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I would say it is equivalent to the map $H^1(G,B) \to H^1(G,B/A)$ being surjective! –  Sasha Aug 27 '10 at 18:32

Sasha's comment is correct. In your case you have the trivial action on $p$-power order cyclic modules. So let me write the map $B \to B/A$ as $B = \mathbb Z/p^n \to \mathbb Z/p^m = B/A,$ where $m \leq n$, and the map is the natural one (reduce a mod $p^n$ class to a mod $p^m$ class).
Now $H^1$ of $G$ against a trivial module is just homs of $G^{ab}$ into this module, so we have to look at $$Hom(G^{ab},\mathbb Z/p^n ) \to Hom(G^{ab},\mathbb Z/p^m).$$ Now $G^{ab}$ is (in your setting) itself an abelian $p$-power order group, so is a product of cyclic $p$-power order groups, so (since $Hom$ from a finite product is the product of the individual $Hom$s) we see that it is enough to consider whether $$Hom(\mathbb Z/p^r,\mathbb Z/p^n) \to Hom(\mathbb Z/p^r,\mathbb Z/p^m)$$ is surjective.
Assuming that $m < n$ (i.e., in the original terms of the problem, that $A$ is non-zero, so that the question is non-trivial), then this map is surjective if and only if $n \leq r.$ (This is not hard to check; see below for a careful explanation.)
Putting this together for all $r$, we get the following: assuming that $A$ is non-zero, that $B$ is cyclic of $p$-power order, and that $G$ is a $p$-group, then the map of $H^2$ is injective precisely when each cyclic direct summand of $G^{ab}$ has order at least that of $B$.
Proof of surjectivity fact: The $Hom$ space $Hom(\mathbb Z/p^r,M)$ is equal to $M[p^r],$ the $p^r$-torsion subgroup of $M$, for any abelian group $M$ (just look at the image of $1$ mod $p^r$). So we have to consider the surjectivity (or non-surjectivity) of $(\mathbb Z/p^n)[p^r] \to (\mathbb Z/p^m)[p^r],$ which is the map $$p^{\max(0,n-r)}\mathbb Z/p^n\mathbb Z \to p^{\max(0,m-r)}\mathbb Z/p^n \mathbb Z.$$ This is surjective if $n - r \leq 0,$ or if $m = n$, but otherwise is not.