Second cohomology group for trivial group action of elementary abelian group of prime-square order on cyclic group of prime-square order

Description of the group
Suppose $$p$$ is a prime number.

We consider here the second cohomology group for trivial group action

$$\! H^2(G,A)$$

where $$G$$ is the elementary abelian group of prime-square order $$E_{p^2} = \mathbb{Z}_p \times \mathbb{Z}_p$$ and $$A$$ is the cyclic group of prime-square order $$\mathbb{Z}_{p^2}$$.

The group is isomorphic to elementary abelian group of prime-cube order $$E_{p^3} = \mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_p$$.

Computation in terms of group cohomology
The cohomology group can be computed as an abstract group using the group cohomology of elementary abelian group of prime-square order, which in turn can be computed using the Kunneth formula for group cohomology combined with the group cohomology of finite cyclic groups.

We explain here the part of the computation based on the group cohomology of elementary abelian group of prime-square order. As per that page, we have:

$$H^2(G;A) = (\operatorname{Ann}_A(p)) \oplus (A/pA)^2$$

Here, $$A/pA$$ is the quotient of $$A$$ by $$pA = \{ px \mid x \in A \}$$ and $$\operatorname{Ann}_A(p) = \{ x \in A \mid px = 0 \}$$.

In our case, $$A = \mathbb{Z}/p^2\mathbb{Z}$$, so we get that both $$A/pA$$ and $$\operatorname{Ann}_A(p)$$ are both isomorphic to $$\mathbb{Z}/p^2\mathbb{Z}$$. Plugging in, we get:

$$H^2(G;A) = \mathbb{Z}/p\mathbb{Z} \oplus (\mathbb{Z}/p\mathbb{Z})^2 = (\mathbb{Z}/p\mathbb{Z})^3$$

which is the elementary abelian group of order $$p^3$$.

Summary
As mentioned earlier, the information here does not apply to the case $$p = 2$$. For the case $$p = 2$$, see second cohomology group for trivial group action of V4 on Z4.