Group cohomology of symmetric groups

This article gives specific information, namely, group cohomology, about a family of groups, namely: symmetric group.
View group cohomology of group families | View other specific information about symmetric group

Particular cases $n$ $n!$ (order of symmetric group) Symmetric group of degree $n$ Group cohomology page
1 1 trivial group group cohomology of trivial group
2 2 cyclic group:Z2 group cohomology of cyclic group:Z2
3 6 symmetric group:S3 group cohomology of symmetric group:S3
4 24 symmetric group:S4 group cohomology of symmetric group:S4
5 120 symmetric group:S5 group cohomology of symmetric group:S5
6 720 symmetric group:S6 group cohomology of symmetric group:S6

Classifying space and corresponding chain complex

The homology and cohomology groups of the symmetric group $S_n$ are the same as the respective homology and cohomology groups of the configuration space of $n$ unordered points in a countable-dimensional real projective space. For more on the topological perspective, see configuration space of unordered points of a countable-dimensional real projective space on the Topospaces wiki.

Homology groups for trivial group action

FACTS TO CHECK AGAINST (homology group for trivial group action):
First homology group: first homology group for trivial group action equals tensor product with abelianization
Second homology group: formula for second homology group for trivial group action in terms of Schur multiplier and abelianization|Hopf's formula for Schur multiplier
General: universal coefficients theorem for group homology|homology group for trivial group action commutes with direct product in second coordinate|Kunneth formula for group homology

Over the integers

Here, "0" for a group is shorthand for the trivial group. $\mathbb{Z}_m$ is shorthand for the finite cyclic group $\mathbb{Z}/m\mathbb{Z}$.

The homology groups eventually stabilize in the following sense: for any fixed $q$, there exists a large enough $n$ (explicit expression for $n$ in terms of $q$ -- around double?) such that $H_q(S_m;\mathbb{Z}) \cong H_q(S_n;\mathbb{Z})$ for all $m \ge n$. The corresponding stable value of homology group is termed the stable homology group of degree $q$ for the symmetric groups. $n$ $n!$ symmetric group $S_n$ of degree $n$ if a finite group with periodic cohomology, period of sequence of homology groups for positive degrees $H_1(S_n;\mathbb{Z})$ $H_2(S_n;\mathbb{Z})$ $H_3(S_n;\mathbb{Z})$ $H_4(S_n;\mathbb{Z})$ $H_5(S_n;\mathbb{Z})$ $H_6(S_n;\mathbb{Z})$ $H_7(S_n;\mathbb{Z})$ $H_8(S_n;\mathbb{Z})$ $H_9(S_n;\mathbb{Z})$ $H_{10}(S_n;\mathbb{Z})$
0 1 trivial group 1 0 0 0 0 0 0 0 0 0 0
1 1 trivial group 1 0 0 0 0 0 0 0 0 0 0
2 2 cyclic group:Z2 2 $\mathbb{Z}_2$ 0 $\mathbb{Z}_2$ 0 $\mathbb{Z}_2$ 0 $\mathbb{Z}_2$ 0 $\mathbb{Z}_2$ 0
3 6 symmetric group:S3 4 $\mathbb{Z}_2$ 0 $\mathbb{Z}_6$ 0 $\mathbb{Z}_2$ 0 $\mathbb{Z}_6$ 0 $\mathbb{Z}_2$ 0
4 24 symmetric group:S4 -- $\mathbb{Z}_2$ $\mathbb{Z}_2$ $\mathbb{Z}_{12} \oplus \mathbb{Z}_2$ $\mathbb{Z}_2$ $(\mathbb{Z}_2)^3$ $(\mathbb{Z}_2)^2$ $\mathbb{Z}_{12} \oplus (\mathbb{Z}_2)^2$ $(\mathbb{Z}_2)^3$ $(\mathbb{Z}_2)^4$ $(\mathbb{Z}_2)^3$
5 120 symmetric group:S5 -- $\mathbb{Z}_2$ $\mathbb{Z}_2$ $\mathbb{Z}_{12} \oplus \mathbb{Z}_2$ $\mathbb{Z}_2$ $(\mathbb{Z}_2)^3$ $(\mathbb{Z}_2)^2$ $\mathbb{Z}_{60} \oplus (\mathbb{Z}_2)^2$ $(\mathbb{Z}_2)^3$ $(\mathbb{Z}_2)^4$ $(\mathbb{Z}_2)^3$
6 720 symmetric group:S6 -- $\mathbb{Z}_2$ $\mathbb{Z}_2$ $\mathbb{Z}_{12} \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ $(\mathbb{Z}_2)^5$ $(\mathbb{Z}_2)^5$ $\mathbb{Z}_{60} \oplus \mathbb{Z}_6 \oplus (\mathbb{Z}_2)^5$ $(\mathbb{Z}_2)^8$
7 5040 symmetric group:S7 -- $\mathbb{Z}_2$ $\mathbb{Z}_2$ $\mathbb{Z}_{12} \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ $(\mathbb{Z}_2)^5$ $(\mathbb{Z}_2)^5$
Stable -- -- -- $\mathbb{Z}_2$ $\mathbb{Z}_2$ $\mathbb{Z}_{12} \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$  ?  ?  ?  ?  ?  ?  ?