# Group (Co)Homology of Symmetric Group

The question concerns the group homology or group cohomology of symmetric groups.

1. 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, $H_k(\text{S}_{n=4},\mathbb{Z})$:

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

1. 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).)

• + 1, Jeremy Rickard, thanks for reminding me. – wonderich Apr 8 '18 at 17:48
• Since symmetric groups are finite groups, its (co)homology is entirely torison (except in $H_0$. – user43326 Apr 8 '18 at 17:50
• Just in case my previous comment was not clear enough: $R/Z$ is injective, so in the universal coefficient theorem's statement, there is no $Ext ^1$ term, so you only get $Hom$ term. However, for any torsion abelian group $A$ you have $Hom(A,R/Z)\cong A$ in a non-canonical way. – user43326 Apr 8 '18 at 18:33
• @ user43326, Thanks so much +1. then, am I correct about my statements in part 2, after all? How about the part 3? [Feel free to write an answer - it will not have to be demanding one.] – wonderich Apr 8 '18 at 19:14
• @user43326 it's false (that $Hom(A,R/Z)$ is isomorphic to $A$ for $A$ torsion abelian). Two counterexamples (among others) (a) if $A=\bigoplus A_n$ with $A_n\neq 0$ finite, then $A$ is countable, while $Hom(A,R/Z)\simeq\prod_n A_n$, which is uncountable. (b) if $A=\mathbf{Z}[1/n]/\mathbf{Z}$, then $Hom(A,R/Z)\simeq\mathbf{Z}_n$, which is torsion-free (and also uncountable). Actually when $A$ is infinite countable, it's always uncountable. – YCor Apr 8 '18 at 22:04

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 $Hom(A,{\mathbb Q}/{\mathbb Z})$.