(co)homology of symmetric groups

Question

Let $S_n=\{\text{bijections }[n]\to[n]\}$ be the n-th symmetric group. Its (co)homology will be understood with trivial action. What are the $\mathbb{Z}$-modules $H_k(S_n;\mathbb{Z})$? Using GAP, we calculate this for small $n$ and $k$: This seems to be an infinite amount of data with no apparent patterns, just the stabilization for $n\geq2k$.

In Stable homology of automorphism groups of free groups (Galatius - 2008) p.2 there is written: "The homology groups $H_k(S_n)$ are completely known" referring to Nakaoka's articles Decomposition Theorem for Homology Groups of Symmetric Groups, Homology of the Infinite Symmetric Group, Note on cohomology algebras of symmetric groups from 1960, 1961, 1962. I haven't found any such table in those articles, or in Cohomology of Finite Groups (Adem, Milgram - 1994). My questions are:

1) Does $H_k(S_n;\mathbb{Q})$ and $H_k(S_n;\mathbb{Z}_p)$ for all prime $p$ determine $H_k(S_n;\mathbb{Z})$?

2) How does the above table look for larger n and k, e.g. what is $H_k(S_{2k};\mathbb{Z})$ for $k=1,...,30$?

3) Is for every prime $p$ and $k\geq1$ the module $\mathbb{Z}_{p^k}$ a direct summand of some $H_k(S_n;\mathbb{Z})$?

4) Does $H_k(S_n;R)\cong H_k(S_{2k};R)$ as $R$-modules for $n>2k$ hold over any ring $R$?

Answer 1

The paradoxical answer is that it is annoying but straightforward to determine $H_k(S_n)$ for particular values of $k$ and $n$, but it is very easy to write down $H_*(\coprod BS_n)$, encompassing all $k$ and all $n$ at once. More generally, if $C$ is the monad on based spaces $X$ determined by any $E_{\infty}$ operad $\mathcal C$, then the Hopf algebra $H_*(CX;\mathbb F_p)$ is an explicitly known functor of the coalgebra $H_*(X;\mathbb F_p)$. Moreover the Bockstein spectral sequence of $CX$ for any prime $p$ is functorially determined by that of $X$, so in principal the integral homology of $CX$ is explicitly determined by the integral homology of $X$. When $X= S^0$, $CX$ is the disjoint union of the classifying spaces $BS_n$, so the homology of all symmetric groups is there as a special case, connected together by multiplicative structure determined by the the evident homomorphisms $S_m\times S_n \longrightarrow S_{m+n}$. User43326 gives a reference for all of this. The answer to 1) and 3) is yes, the answer to 2) is that it is boring but implicit how to write down a table such as yours for small values of $n$ and $k$, it is just not especially interesting to do so, or so it seems to me. It can be left as an exercise to check whether or not 4) is true.

Answer 2

The answer to the question 1) is "yes". The classifying space of the symmetric group is of finite type, so its integral homology is determined by its rational homology and $p$-local homology for all $p$'s. Now, mod $p$ homology doesn't really determine the $p$-local homology, but we know completely the bockstein spectral sequence (see http://www.math.uchicago.edu/~may/BOOKS/homo_iter.pdf Chapter 1, Theorem 4.13, so we can get the $p$-local homology. Furthermore, since the symmetric group is finite, its rational homology is trivial.

The answer to the question 4) is also yes (and this time, it is a real yes), and a good reference is http://www.math.uchicago.edu/~may/BOOKS/homo_iter.pdf You can read off mod $p$ homology of $\Sigma _n$ from that of $CS^0$, and this is treated in chapter 1, section 5.

Hopefully you can find the answers to other questions in the references above.

Answer 3

Here are some comments, including an answer to (3). Firstly, if you want an actual explicit computation of the mod-2 cohomology of symmetric groups $S_n$ for as large an $n$ as possible, you should look at M. Feshbach's `The mod-2 cohomology rings of the symmetric groups and invariants' Topology volume 41 (2002) 57-84. This contains explicit computations of the cohomology rings $H^*(S_n;\mathbb{F}_2)$ for $n\leq 16$, including correcting a minor error in a calculation in the book of A Adem and J Milgram.

The advantage of cohomology is that it is a ring, also that the cohomology of a finite group with coefficients in either $\mathbb{Z}$ or a field is a finitely presented algebra over the coefficients.

Here is a non-constructive solution to (3). By the universal coefficient theorem, it suffices to show the same thing for cohomology. Now let $n$ be sufficiently large that the symmetric group $S_n$ contains a cyclic subgroup of order $p^k$; of course the least such $n$ is $n=p^k$. For any such $n$, there will be elements of order $p^k$ in $H^j(S_n;\mathbb{Z})$ for infinitely many values of $j$. To see this one uses the Evens-Venkov theorem.

For any finite group $G$ and any subgroup $H$, the map from $H^*(G;\mathbb{Z})$ to $H^*(H;\mathbb{Z})$ makes the ring $H^*(H;\mathbb{Z})$ into a module for the ring $H^*(G;\mathbb{Z})$. The Evens-Venkov theorem tells us that $H^*(H;\mathbb{Z})$ is finitely generated as an $H^*(G;\mathbb{Z})$-module.

Now to apply this. The cohomology ring of the cyclic group of order $p^k$ is isomorphic to a polynomial ring $\mathbb{Z}[c]/(p^kc)$, where $c$ is a generator for $H^2$. If $R$ is a (graded) subring of this ring such that the whole ring is a finitely generated $R$-module, then $R$ contains $c^m$ for some $m$, and hence $H^{2mj}(S_n;\mathbb{Z})$ contains an element of order $p^k$ for every $j$. The universal coefficient theorem then tells you that $H_{2mj-1}(S_n;\mathbb{Z})$ contains an element of order $p^k$ for all $j$.

You asked about summands of order exactly $p^k$. In fact, $H^*(S_n;\mathbb{Z})$ will contain these whenever $n\geq p^k$, although I can't provide a quick argument. A quicker thing to see is that for $p^k \leq n < p^{k+1}$, the exponent of the $p$-local cohomology of $S_n$ is exactly $p^k$. I'll give the argument just for $n=p^k$. There is a subgroup of $S_{p^k}$ isomorphic to the direct product of $p$ copies of $S_{p^{k-1}}$. Furthermore, the index of this subgroup is divisible by $p$ but not divisible by $p^2$.

For any finite group $G$ and subgroup $H$, there is a transfer map in cohomology $H^*(H;\mathbb{Z})\rightarrow H^*(G;\mathbb{Z})$ with the property that the composite map from $H^*(G;\mathbb{Z})$ to itself given by first mapping to $H^*(H;\mathbb{Z})$ and then transferring back up is equal to multiplication by the index $|G:H|$.

Going back to the symmetric group, we know by induction on $k$ that the exponent of the $p$-part of $H^*(S_{p^{k-1}};\mathbb{Z})$ is $p^{k-1}$, and by the Kunneth formula the same holds true for the direct product of $p$ copies of this group. The restriction from $S_{p^k}$ down to this subgroup, followed by the transfer map back up is, up to units, multiplication by $p$ on the $p$-local cohomology $H^*(S_{p^k};\mathbb{Z}_{(p)})$. Hence the exponent of this group is at most $p^k$. Combined with the Evens-Venkov lower bound this gives the claim.