Second cohomology group for trivial group action of S4 on Z2

Description of the group
This article describes the second cohomology group for trivial group action

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

where $$G$$ is symmetric group:S4 (the symmetric group on a set of size four) and $$A$$ is cyclic group:Z2. $$G$$ itself has order 24 and $$A$$ has order 2.

The cohomology group $$H^2(G;A)$$ is isomorphic to the Klein four-group.

Computation of the group
See group cohomology of symmetric group:S4. It is clear that we have:

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

In the case that $$A$$ is cyclic of order two, both the groups $$A/2A$$ and $$\operatorname{Ann}_A(2)$$ are cyclic of order two, so their direct sum is the Klein four-group.

Elements
Note that in all cases, the base of the group extension is characteristic in the whole group because it is precisely the center of the whole group. This is because the quotient group is a centerless group and hence cannot partake of any of the center.

General background
We know from the general theory that there is a natural short exact sequence:

$$0 \to \operatorname{Ext}^1_{\mathbb{Z}}(G^{\operatorname{ab}},A) \to H^2(G;A) \to \operatorname{Hom}(H_2(G;\mathbb{Z}),A) \to 0$$

where $$G^{\operatorname{ab}}$$ is the abelianization of $$G$$ (i.e., the quotient of $$G$$ by its derived subgroup) and $$H_2(G;\mathbb{Z})$$ is the Schur multiplier of $$G$$.

The image of $$\operatorname{Ext}^1$$ in $$H^2$$ comprises precisely the extensions where the derived subgroup of $$G$$ (in this case A4 in S4) splits as a direct product and the quotient group sits in an abelian fashion on top of the center. We also know, again from the general theory, that the short exact sequence above splits. However, there is no generic natural choice of the splitting. There need not in general be a natural or even an automorphism-invariant choice of splitting.

In this case
In this case, the image of $$\operatorname{Ext}^1(G^{\operatorname{ab}},A)$$ is cyclic group:Z2 (because $$G^{\operatorname{ab}}$$ itself is isomorphic to cyclic group:Z2) and comprises the extensions direct product of S4 and Z2 and special linear group:SL(2,Z4).

The quotient group, namely $$\operatorname{Hom}(H_2(G;\mathbb{Z}),A)$$, is also isomorphic to cyclic group:Z2, because $$H_2(G;\mathbb{Z})$$ is isomorphic to cyclic group:Z2.

The non-identity coset of the image of $$\operatorname{Ext}^1$$ in $$H^2$$ comprises the two stem extensions binary octahedral group and general linear group:GL(2,3).

There are two different choices of complement to this that we can use to split the sequence. The two ways of splitting are both automorphism-invariant and are depicted below. The rows represent cosets of the image of $$\operatorname{Ext}^1$$ and the columns represent cosets of the chosen complement. The first row and first column respectively represent the subgroups themselves.

First way of splitting:

Second way of splitting:

More on the mapping from $$\operatorname{Ext}^1$$
Below is an explicit description of the mapping from $$\operatorname{Ext}^1(G^{\operatorname{ab}},A)$$ to $$H^2(G;A)$$, in terms of the original extension for cyclic group:Z2 (the abelianization of symmetric group:S4) on top of cyclic group:Z2 and what the new extension looks like for symmetric group:S4 on top of cyclic group:Z2.