TITLE: Cohomology ring of $H^*(A_5;\mathbb Z_2)$?
QUESTION [14 upvotes]: What is the cohomology ring of $H^*(A_5;\mathbb Z_2)$? Here $A_5$ is the alternating group on $5$ letters. 
I am comfortable with the Lyndon-Hochschild-Serre spectral sequence, and understand how to use it to compute the cohomology of $A_4$. However, because $A_5$ is simple, there is no extension $1 \to G \to A_5 \to H \to 1$, so I'm not sure how to use this.

REPLY [5 votes]: There is a much easier answer that I missed on a first pass, but uses some technical theory as opposed to just doing a spectral sequence. 
Let $G$ be a group and $P$ be the $p$-Sylow subgroup. If $P$ is abelian, then the map $$H^*(G;\Bbb F_p) \to H^*(P;\Bbb F_p)^{N_G(P)/P}$$ given by restriction is an isomorphism; here the exponent means we take fixed points under the natural action of $N_G(P)/P$ by conjugation (we may quotient by $P$ because it acts trivially: $P$ is abelian). The proof uses the double coset formula for the transfer operator; you may find such a proof in Theorem 6.8 of Adem and Milgram's book on group cohomology.
In particular, we have $$H^*(A_5; \Bbb F_2) = H^*(V_4; \Bbb F_2)^{\Bbb Z/3}$$
$$H^*(A_5; \Bbb F_3) = H^*(\Bbb Z/3; \Bbb F_3)^{\Bbb Z/2}$$
$$H^*(A_5; \Bbb F_5) = H^*(\Bbb Z/5; \Bbb F_5)^{\Bbb Z/2}$$
I'll let you calculate the actions of $N_G(P)/P$ on $P$ in each case.
The most interesting is the top. I will cite for you the full computation from Adem-Milgram, Theorem 1.3: $$H^*(A_4; \Bbb F_2) = \Bbb F_2[w_2, w_3, k]/(k^2 + kw_3 + w_3^2 + w_2^3).$$
The other two are quick calculations, and give $$H^*(A_5; \Bbb F_3) = \Bbb F_3[k, p_1]/(k^2)$$ and $$H^*(A_5; \Bbb F_5) = \Bbb F_5[k, p_1]/(k^2)$$ as expected.