Let $S\_3$ be the symmetric group of order $3$. What is the cohomology ring


$$
H^\*(S\_3;\mathbb{Z})?$$


My attempt: I want to use mathematical induction on $n$ for $S\_n$.
For $n=1$, $S\_1$ is trivial. Hence 
$$
H^\*(S\_1;\mathbb{Z})=\mathbb{Z}.$$
 For $n=2$, $BS\_2=\mathbb{R}P^\infty$. Hence 
$$
H^\*(S\_2;\mathbb{Z})=H^\*(\mathbb{R}P^\infty;\mathbb{Z})=\mathbb{Z}\_2[\alpha],$$
where $|\alpha|=2$. But I do not know how to continue.



cohomology ring of symmetric group of order $3$
Your ring for $S\_2=\mathbb{Z}\_2$ is wrong, because $H^0=\mathbb{Z}$. It should be $\mathbb{Z}[\alpha]/(2\alpha)$. 


The cohomology groups of $S\_3=\mathbb{Z}\_3\rtimes \mathbb{Z}\_2$ are computed using the $p$-primary decomposition, $H^i(S\_3)=H^i(S^3)\_{(2)}\oplus H^i(S^3)\_{(3)}$. I provided this long computation as an answer on Math StackExchange, [here](https://math.stackexchange.com/questions/185081/calculating-the-group-co-homology-of-the-symmetric-group-s-3-with-integer-coef/210345#210345). It is $\mathbb{Z}\_2$ in degrees 2 mod 4, and $\mathbb{Z}\_6$ in nonzero degrees 0 mod 4, and 0 in odd degrees, and $\mathbb{Z}$ in degree 0.


This group has periodic cohomology of period 4, which means the generator $x\in H^4(S\_3)=\mathbb{Z}\_6$ induces isomorphisms $H^i\cong H^{i+4}$ via cup product. And the generator $y\in H^2(S\_3)=\mathbb{Z}\_2$ satisfies $y^2=3x$ because $H^4(S^3)\_{(3)}=\mathbb{Z}\_3$ while $H^2(S^3)\_{(3)}=0$.


Thus $H^\ast(S\_3)\cong\mathbb{Z}[x,y]/(6x,2y,y^2-3x)$ where $|x|=4$ and $|y|=2$.




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**Alternative:** A group $G$ with periodic cohomology has its cohomology ring isomorphic to the associated graded ring of the representation ring $R(G)$. In general, $H^\ast(G)$ and $R(G)$ are related by a spectral sequence, and this was shown by Atiyah in *"Characters and Cohomology of Finite Groups"*. Section 13 computes the generators explicitly for $S\_3$, but he didn't completely write down the ring structure: Here $R\_{2k-1}(S\_3)=R\_{2k}(S\_3)$ and $R\_2(S\_3)=\lbrace \alpha,\beta\rbrace$ and $R\_4(S\_3)=\lbrace 2\alpha,\alpha+\beta\rbrace$, such that $\alpha^2=\alpha\beta=2\alpha$ and $\beta^2=3\beta-\alpha$. Thus $R\_6(S\_3)=R\_2(S\_3)R\_4(S\_3)=\lbrace 4\alpha,2\alpha+\beta^2\rbrace$, so that $\alpha\,\text{mod}\,R\_4(S\_3)$ generates $H^2(S\_3)$ and $\alpha+\beta\,\text{mod}\,R\_6(S\_3)$ generates $H^4(S\_3)$. These generators are related by $3(\alpha+\beta)=4\alpha+\beta^2\equiv 2\alpha=\alpha^2\,\text{mod}\,R\_6(S\_3)$, so we get agreement with the ring that I gave.


