I have been trying for a while to make sense of Ex V.3.5 & Ex III.10.1 in Brown's book 'Co-homology of Groups': Calculate the Co-homology of $S_3$ with co-efficients in $\mathbb{Z}$, possibly using cup products.

Ex III.10.1 says that it is possible to calculate this using III.10.3: The p-primary part of $H^n(S_3, M)$ is isomorphic to the $S_3$-invariant elements of $H^n(H, M)$ where $H$ is a Sylow $p$-subgroup of $G$.

The Sylow $p$-subgroups of $S_3$ are a copy of $\mathbb{Z}_3$ and $\mathbb{Z}_2$, so the first step I suppose is to compute the $S_3$-invariant elements of $H^2(\mathbb{Z}_3, \mathbb{Z})$. The only method I know to compute these is using a resolution $F_*$ of $\mathbb{Z}$ by $\mathbb{Z}[S_3]$ modules: finding a generator of $H^2(\mathbb{Z}_3, \mathbb{Z})$ in the chain complex $\mathcal{Hom}_{\mathbb{Z}[\mathbb{Z}_3]}(F_*, \mathbb{Z})$ and watching what happens to it under the action of $S_3$. At this point I am completely stuck, I cannot think of a way to explicitly find a generator of $H^2(\mathbb{Z}_3, \mathbb{Z})$ in this chain complex.

So I am wondering, have I started going about this the right way? Perhaps there is an easier chain complex in which I can compute the $S_3$-invariant elements, or a way to compute what such a generator of $H^2(\mathbb{Z}_3, \mathbb{Z})$?

Thanks for any help that anyone can give.

  • $\begingroup$ @DylanMoreland Opps, yes I am talking about Brown's book, I've added that. Thanks. $\endgroup$ Commented Aug 21, 2012 at 16:09
  • $\begingroup$ This question is related to mathoverflow.net/questions/92567/… $\endgroup$ Commented Aug 22, 2012 at 9:23
  • $\begingroup$ The symmetric group $S_3$ is the dihedral group $D_3$. $\endgroup$
    – Leo
    Commented Sep 11, 2014 at 14:13

1 Answer 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


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