Calculating the group co-homology of the symmetric group $S_3$ with integer coefficients.

Solution 1:

Here are my solutions from when I was studied under Ken Brown as an undergraduate:

V.3.5: From the semi-direct product representation $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{Z}_3$ by conjugation, $H^*(S_3)=H^*(S_3)_{(2)}\oplus H^*(S_3)_{(3)}\cong H^*(S_3)_{(2)}\oplus H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ by Theorem III.10.3, with $H^*(S_3)_{(2)}$ isomorphic to the set of $S_3$-invariant elements of $H^*(\mathbb{Z}_2)$. Exercise III.10.1 showed that $H^*(S_3)_{(2)}\cong \mathbb{Z}_2$, so it suffices to compute $H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ where we know that $\mathbb{Z}_2$ acts by conjugation on $\mathbb{Z}_3$ ($1\mapsto 1$, $x\mapsto x^2$, $x^2\mapsto x$). But this action can be considered as the endomorphism $\alpha(2)$ from Exercise V.3.4 since $(1)^2=1$ and $(x)^2=x^2$ and $(x^2)^2=x^4=x$, and that exercise implies that the induced map on the $(2i)^{th}$-cohomology is multiplication by $2^i$ [we know that the cohomology is trivial in odd dimensions]. Now $2^1\equiv 2\,\text{mod}3$ and $2^2\equiv 1\,\text{mod}3$, so by multiplying both of those statements by $2$ repeatedly we see that $2^i\equiv 1\,\text{mod}3$ for $i$ even and $2^i\equiv 2\,\text{mod}3$ for $i$ odd. Thus the largest $\mathbb{Z}_2$-submodule of $H^{2i}(\mathbb{Z}_3)\cong\mathbb{Z}_3$ on which $\mathbb{Z}_2$ acts trivially is $\mathbb{Z}_3$ (itself) for $i$ even, and is $0$ for $i$ odd. It now follows that the integral cohomology $H^*(S_3)$ is the same as that which was deduced in Exercise III.10.1, namely, it is $\mathbb{Z}_2$ in the $2\,\text{mod}4$ dimensions and is $\mathbb{Z}_6$ in the nonzero $0\,\text{mod}4$ dimensions and is $0$ otherwise (besides the $0^{th}$-dimension in which it is $\mathbb{Z}$).

III.10.1: From the semi-direct product representation $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{Z}_3$ by conjugation, $H^*(S_3)=H^*(S_3)_{(2)}\oplus H^*(S_3)_{(3)}\cong H^*(S_3)_{(2)}\oplus H^*(\mathbb{Z}_3)^{\mathbb{Z}_2}$ by Theorem III.10.3. Now $S_3$ is the unique nonabelian group of order 6, so $D_6\cong S_3$ and we can use a result (given below as "additional exercise") which implies that the $\mathbb{Z}_2$-action on $H_{2i-1}(\mathbb{Z}_3)\cong H^{2i}(\mathbb{Z}_3)$ is multiplication by $(-1)^i$. Thus $H^n(\mathbb{Z}_3)^{\mathbb{Z}_2}$ is isomorphic to $\mathbb{Z}_3$ for $n=2i$ where $i$ is even, and is trivial for $n$ odd and $n = 2i$ where $i$ is odd. Taking any Sylow $2$-subgroup $H\cong \mathbb{Z}_2$, Theorem III.10.3 states that $H^*(S_3)_{(2)}$ is isomorphic to the set of $S_3$-invariant elements of $H^*(H)$. In particular we have the monomorphism $H^{2i-1}(S_3)_{(2)}\hookrightarrow H^{2i-1}(H)=0$, so $H^{2i-1}(S_3)_{(2)}=0$. An $S_3$-invariant element $z\in H^{2i}(H)\cong\mathbb{Z}_2$ must satisfy the equation $\text{res}^H_Kz=\text{res}^{gHg^{-1}}_Kgz$, where $K$ denotes $H\cap gHg^{-1}$. If $g\in H$ then $gHg^{-1}=H$ and the above condition is trivially satisfied for all $z$ ($hz=z$ by Proposition III.8.1). If $g\notin H$ then $K=\lbrace 1\rbrace$ because $H$ is not normal in $S_3$ and only contains two elements, so the intersection must only contain the trivial element. But then the image of both restriction maps is zero, so the condition is satisfied for all $z$; thus $H^{2i}(S_3)_{(2)}=\mathbb{Z}_2$. Alternatively, a theorem of Richard Swan states that if $G$ is a finite group such that $\text{Syl}_p(G)$ is abelian and $M$ is a trivial $G$-module, then $\text{Im}(\text{res}^G_{\text{Syl}_p(G)})=H^*(\text{Syl}_p(G),M)^{N_G(\text{Syl}_p(G))}$. It is a fact that $N_{S_3}(\mathbb{Z}_2)=\mathbb{Z}_2$, so taking $G=S_3$ and $H=\text{Syl}_2(S_3)\cong\mathbb{Z}_2$ and $M=\mathbb{Z}$ we have $\text{Im}(\text{res}^{S_3}_H)=(\mathbb{Z}_2)^{\mathbb{Z}_2}=\mathbb{Z}_2$ in the even-dimensional case. Since any invariant is in the image of the above restriction map (by Theorem III.10.3), the result $H^{2i}(S_3)_{(2)}=\mathbb{Z}_2$ follows.

Additional Exercise: The cyclic group $C_m$ is a normal subgroup of the dihedral group $D_m=C_m\rtimes C_2$ (of symmetries of the regular $m$-gon). There is a $C_2$-action on $C_m=\langle \sigma\rangle$ given by $\sigma\mapsto \sigma^{-1}$. Determine the action of $C_2$ on the homology $H_{2i-1}(C_m,\mathbb{Z})$, noting that there is an element $g\in D_m$ such that $g\sigma g^{-1}=\sigma^{-1}$.
Solution: Letting $c(g):C_m\rightarrow C_m$ be conjugation by $g$, we can apply Corollary III.8.2 to obtain the induced action of $D_m/C_m\cong C_2$ on $H_*(C_m,\mathbb{Z})$ given by $z\mapsto c(g)_*z$. It suffices to compute $c(g)_*$ on the chain level, using the periodic free resolution $P$ of $C_m$, and using the trivial action on $\mathbb{Z}$. Using the condition $\tau(hx)=[c(g)](h)\tau(x)=h^{-1}\tau(x)$ on the chain map $\tau:P\rightarrow P$ (for $h\in C_m$), we claim that $\tau_{2i-1}(x)=\tau_{2i}(x)=(-1)^i\sigma^{m-i}x$ for $i\in\mathbb{N}$ and $\tau_0(x)=x$. Assuming this claim holds, the chain map $P\otimes_{C_m}\mathbb{Z}\rightarrow P\otimes_{C_m}\mathbb{Z}$ [in odd dimensions] is given by $x\otimes y\mapsto (-1)^i\sigma^{m-i}x\otimes gy=(-1)^i\sigma^{m-i}x\otimes y=(-1)^ix\otimes \sigma^{i-m}y=(-1)^ix\otimes y$, and so $c(g)_*$ [hence the $C_2$-action] is multiplication by $(-1)^i$ on $H_{2i-1}(C_m,\mathbb{Z})$. It suffices to prove the claim. Seeing that $N\tau_{2i}(1)=\tau_{2i-1}(N)=N\tau_{2i-1}(1)$ where $N$ is the norm element, we can restrict our attention to $\tau_{2i-1}$ and use induction on $i$ since $(\sigma-1)\tau_1(1)=(\sigma-1)(-\sigma^{m-1})=\sigma^{m-1}-1=\sigma^{-1}-1=\tau_0(\sigma-1)$. This chain map must satisfy commutativity $(\sigma-1)\tau_{2i-1}(1)=\tau_{2(i-1)}(\sigma-1)$, and this is indeed the case because $(\sigma-1)\tau_{2i-1}(1)=(-1)^i(\sigma^{m-i+1}-\sigma^{m-i})$ and $\tau_{2(i-1)}(\sigma-1)=(\sigma^{-1}-1)(-1)^{i-1}\sigma^{m-i+1}=(-1)^i(\sigma^{m-i+1}-\sigma^{m-i})$.

Recap: $H^n(S_3)\cong$
$\mathbb{Z}$ for $n=0$,
$\mathbb{Z}_6$ for $n\equiv 0\;\text{mod}\,4$ with $n\ne 0$,
$\mathbb{Z}_2$ for $n\equiv 2\;\text{mod}\,4$,
$0$ otherwise