TITLE: Group (Co)Homology of Symmetric Group
QUESTION [6 upvotes]: The question concerns the group homology or group cohomology of symmetric groups.

The entries in groupprops.subwiki.org and in this MO post show the results for the symmetric group S$_4$.

groupprops.subwiki.org, $H_q(\text{S}_4,\mathbb{Z})$:

this MO post, $H_k(\text{S}_{n=4},\mathbb{Z})$:


If the groupprops.subwiki.org has the correct result, is it correct to say that the torsion parts of Group (Co)Homology of Symmetric Group are related by:

$$H^q(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_q(\text{S}_4,\mathbb{Z})?$$
Thus,
$$H^1(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_1(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2?$$
$$H^2(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_2(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2?$$
$$H^3(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_3(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2\oplus \mathbb{Z}_4\oplus \mathbb{Z}_3?$$
$$H^4(\text{S}_4,\mathbb{R}/\mathbb{Z})=H_4(\text{S}_4,\mathbb{Z})=\mathbb{Z}_2?$$

More generally, do we have a precise relation between the group cohomology and group homology of different coefficients:

$$H^q(G,\mathbb{R}/\mathbb{Z}) \text { and } H_p(G,\mathbb{Z}),$$
say for any finite group $G$, and here in particular $\text{S}_4$? (Here $\mathbb{R}/\mathbb{Z}=S^1=$U(1).)

REPLY [5 votes]: First of all, let $X$ be a space (with usual nice properties).  Then 
as ${\mathbb Q}/{\mathbb Z}$ is injective, the universal coefficient theorem for cohomology gives an isomorphism
$$Hom (H_*(X,{\mathbb Z}) ,{\mathbb Q}/{\mathbb Z})
\cong H^*(X,{\mathbb Q}/{\mathbb Z}). $$
Now, when $X$ is the classifying space of a finite group $BG$, in positive dimension, $A=H_*(X,{\mathbb Z})$ is finite abelian group, thus there is non-canonical isomorphism between $A$ and $Hom(A,{\mathbb Q}/{\mathbb Z})$.
Thus the answer to the third question (which includes the second one) is "yes".