Formula for second cohomology group for trivial group action for abelian groups of prime power order

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Statement

General case

Suppose p is a prime number and G and A are finite abelian p-groups. Denote by H^2(G;A) the second cohomology group for trivial group action of G on A. Then, H^2(G;A) is also a finite p-group and is given as follows:

Suppose:

\! A \cong \bigoplus_{i=1}^r \mathbb{Z}/p^{a_i}\mathbb{Z}

and

\! G \cong \bigoplus_{j=1}^s \mathbb{Z}/p^{b_j}\mathbb{Z}

Then:

\! H^2(G;A) \cong \operatorname{Ext}^1(G;A) \oplus \operatorname{Hom}(\bigwedge^2G,A)

where:

\! \operatorname{Ext}^1(G;A) \cong \bigoplus_{1 \le i \le r, 1 \le j \le s} \mathbb{Z}/p^{\min\{a_i,b_j\}}\mathbb{Z}

and

\! \operatorname{Hom}(\bigwedge^2G,A)\cong \bigoplus_{1 \le i \le r, 1 \le j < k \le s} \mathbb{Z}/p^{\min\{a_i,b_j,b_k\}}\mathbb{Z}

Combining, we get that:

\! H^2(G;A) \cong \bigoplus_{1 \le i \le r, 1 \le j \le s} \mathbb{Z}/p^{\min\{a_i,b_j\}}\mathbb{Z} \oplus \bigoplus_{1 \le i \le r, 1 \le j < k \le s} \mathbb{Z}/p^{\min\{a_i,b_j,b_k\}}\mathbb{Z}

Thus, we get the following formula for the prime-base logarithm of order:

\! \log_p|H^2(G;A)| = \sum_{1 \le i \le r, 1 \le j \le s} \min\{a_i,b_j\} + \sum_{1 \le i \le r, 1 \le j < k \le s} \min\{a_i,b_j,b_k\}

Case where the base group is elementary abelian

In the special case that A is an elementary abelian group of prime power order, we get A \cong (\mathbb{Z}/p\mathbb{Z})^r and hence:

\! \operatorname{Ext}^1(G;A) \cong (\mathbb{Z}/p\mathbb{Z})^{rs}

and

\! \operatorname{Hom}(\bigwedge^2G,A)\cong (\mathbb{Z}/p\mathbb{Z})^{rs(s-1)/2}

So that overall:

\! H^2(G;A) \cong (\mathbb{Z}/p\mathbb{Z})^{rs(s+1)/2}

and thus:

\! \log_p|H^2(G;A)| = \frac{rs(s+1)}{2}

Related facts

Facts used

  1. Formula for second cohomology group for trivial group action in terms of Schur multiplier and abelianization