# mod p cohomology ring of alternating groups

Let $A_n$ be the alternating group of $\{1,2,\cdots,n\}$.

(1). What is the cohomology ring $$H^*(A_4;\mathbb{Z}/3)$$ and its Steenrod operation $P^i$'s?

(2). Are there general results about the cohomology ring $$H^*(A_{p+1};\mathbb{Z}/p)$$ for general primes $p\geq 3$?

(3). What is the cohomology ring $$H^*(A_n;\mathbb{Z}_2)$$ for $n\geq 4$ (we can impose conditions on $n$)?

Any references?

• For odd prime $p$, he $p$-sylow subgroup of $A_{p+1}$ is $Z/p$, so they are a direct summand of $H^*(BZ/p.Z/p)$. It shouldn't be too hard to determine. – user43326 Oct 4 '15 at 6:34
• Alternating groups have homological stability, and the stable homology is computable via scanning, eg arxiv.org/abs/1306.6896. – skupers Oct 4 '15 at 15:05
• @skupers OP is asking for the homology of individual $A_m$. – user43326 Oct 5 '15 at 8:05

An easier way for the first case

Consider $P$, a $3$-Sylow subgroup of $A_4$. For example, take the cyclic group generated by the cyclic permutation (123). It is self normalizing, so the double coset formula for the compositions $BP\rightarrow BA_4 \stackrel{tr}{\rightarrow}BP$ where $tr$ denotes the transfer, reduces to the identity. Therefore $BP$ and $BA_4$ are homotopy equivalent at the prime 3.

For the second case, let's start with the review of the well-known computation of $H^*(B\Sigma _p;Z/p)$. Again as in the above, denote by $P$ the cyclic group generated by the cyclic permutation $(12\cdots p)$. Denote by $N$ the group of affine automorphisms of $Z/p$, considered as a subgroup of the permutation of $\{1,\ldots ,p\}$. Then $N$ is the normalizer of $P$ in $\Sigma _p$, with $N/P\cong Gl_1(Z/p)$ with the canonical action on $Z/p$. Now the double coset formula for the compositions $BP\stackrel{Bi}{\rightarrow} B\Sigma _p \stackrel{tr}{\rightarrow}BP$ says that in mod $p$ cohomology, this composition is the sum of maps induced by the multiplication by $s\in Gl_1(Z/p)$. In other words, we have, $$Bi^*tr^*(\beta ^{\epsilon}x^i)=\Sigma _{s\in Gl_1(Z/p)}\beta ^{\epsilon}s^ix^i\in H^*(BZ/p,Z/p)\cong Z/p[x]\otimes \Lambda _{Z/p}(\beta x).$$ The little Theorem of Fermat then shows that this is identity if $p-1\mid i$, $0$ otherwise. Therefore we get $$H^*(B\Sigma _p,Z/p)\cong Z/p[x^{p-1}]\otimes \Lambda _{Z/p}(\beta x^{p-1})\subset H^*(BZ/p,Z/p).$$

Now, if we replace $\Sigma _p$ with $A_{p+1}$, what changes is the normalizer. It is easy to see that $$N_{A_{p+1}}(P)=A_{p+1}\cap \Sigma _p.$$ We also see that the generator of $Gl_1(Z/p)$ acts on $P$ as a cyclic permutation of length $p-1$, thus its an odd permutation. This means in the appropriate double coset formula, the only half of $s's$ get involved (squares in $Gl_1(Z/p)$), and we arrive at the conclusion $$H^*(BA _{p+1},Z/p)\cong Z/p[x^\frac{p-1}{2}]\otimes \Lambda _{Z/p}(\beta x^\frac{p-1}{2})\subset H^*(BZ/p,Z/p).$$ When $p=3$, this reduces to the first case treated in the above.

Regarding your first question, you have a group extension $$1\rightarrow (\mathbb{Z}/2)^2\rightarrow A_4\rightarrow \mathbb{Z}/3\rightarrow 1.$$ Then we can use the Lyndon-Hochschild-Serre spectral sequence $$E_2^{p,q}=H^p(\mathbb{Z}/3,H^q((\mathbb{Z}/2)^2,\mathbb{Z}/3))\implies H^{p+q}(A_4,\mathbb{Z}/3)$$ a priori the $E_2$-page involves local coefficient cohomology. But what you get, by triviality of $H^q((\mathbb{Z}/2)^2,\mathbb{Z}/3)$ is that the projection map $A_4\rightarrow \mathbb{Z}/3$ induces an isomorphism of unstable $\mathcal{A}_3$-algebra: $$H^{*}(A_4,\mathbb{Z}/3)\cong H^{*}(\mathbb{Z}/3,\mathbb{Z}/3).$$ And $H^*(\mathbb{Z}/3,\mathbb{Z}/3)$ is isomorphic to the tensor algebra $\Lambda(x)\otimes \mathbb{Z}/3[y]$ where $x$ of degree 1 and $y$ of degree $2$ moreover $y=\beta(x)$.

And you can also use the same spectral sequence to compute $H^*(A_4,\mathbb{Z}/2)$. What you get is an isomorphism of algebras $$H^*(A_4,\mathbb{Z}/2)\cong H^0(\mathbb{Z}/3,H^*((\mathbb{Z}/2)^2,\mathbb{Z}/2))\cong \mathbb{Z}/2[u,v]^{\mathbb{Z}/3}$$ where $u$ and $v$ are of degree 1.

EDIT: If you want to compute the cohomology algebra of $H^*(\mathbb{Z}/3,\mathbb{Z}/3)$. You can use the fact that $$B\mathbb{Z}/3\simeq L(\infty,3)$$ where $L(\infty,3)=\bigcup_n L(n,3)$ is a limit of Lens spaces. And then use the fibration $$S^1\rightarrow L(\infty,3)\rightarrow \mathbb{C}P^{\infty}.$$ Playing with the spectral sequence of this fibration you get the claimed results: $$H^*(\mathbb{Z}/3,\mathbb{Z}/3)\cong H^*(S^1,\mathbb{Z}/3)\otimes H^*(\mathbb{C}P^{\infty},\mathbb{Z}/3)\cong \Lambda(x)\otimes \mathbb{Z}/3[y].$$ In order to compute $\beta(x)$,just look at this spectral sequence over the integers and observe that the differential $$E_2^{0,1}=\mathbb{Z}\langle x \rangle\rightarrow E_2^{2,0}=\mathbb{Z}\langle y \rangle$$ is such that $dx=3y$. Now the action of $\mathcal{A}_3$ is completely determined by the Cartan relation and the fact that:

• $P^k(x)$ when $k>0$ by unstability,
• $P^1(y)=y^3$ (because $y$ is of degree 2) and $P^k(y)=0$ when $k>1$.