cohomology ring of symmetric group of order $3$ 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.
 A: 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. 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 $|y|=2$.

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.
