Cohomology groups of $K(\mathbb{Z}_p \times \mathbb{Z}_p,1)$

Question

I have a question regarding the cohomology groups of the Eilenberg-MacLane space $K(\mathbb{Z}_p \times \mathbb{Z}_p,1)$. For $n$ > $2$, is there a way to show that $H^n(K(\mathbb{Z}_p \times \mathbb{Z}_p,1);\mathbb{Z}_p)$ is not equal to $0$ or $\mathbb{Z}_p$? I came across this while attempting problem 4.2.28 of Hatcher's Algebraic Topology.

I have heard that, in general, the cohomology groups of $K(G,1)$ are isomorphic to the group cohomologies $H^*(G,\mathbb{Z})$, but I cannot find a proof of this in the Hatcher book. Is there an easy proof of this?

Answer 1

An example of a $K(\mathbb{Z}_p, 1)$ is the infinite lens space $S^{\infty}/\mathbb{Z}_p$, so $S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p$ is an example of a $K(\mathbb{Z}_p\times\mathbb{Z}_p, 1)$. Recall that Eilenberg-MacLane spaces are unique up to homotopy equivalence, so it is enough to calculate the cohomology of $S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p$.

When $p = 2$, the infinite lens space $S^{\infty}/\mathbb{Z}_p$ is infinite real projective space $\mathbb{RP}^{\infty}$. Exactly as in the $p = 2$ case, $S^{\infty}/\mathbb{Z}_p$ has a simple CW complex structure consisting of a single cell in each dimension (see Example $2.43$ of Hatcher) which gives rise to a chain complex

$$0 \to \mathbb{Z} \xleftarrow{0} \mathbb{Z} \xleftarrow{p} \mathbb{Z} \xleftarrow{0} \mathbb{Z} \xleftarrow{p} \mathbb{Z} \xleftarrow{0} \dots$$

If we use $\mathbb{Z}_p$ coefficients instead of integers, we obtain the chain complex

$$0 \to \mathbb{Z}_p \xleftarrow{0} \mathbb{Z}_p \xleftarrow{0} \mathbb{Z}_p \xleftarrow{0} \mathbb{Z}_p \xleftarrow{0} \mathbb{Z}_p \xleftarrow{0} \dots$$

The corresponding cochain complex (using $\mathbb{Z}_p$ coefficients) is

$$0 \to \mathbb{Z}_p \xrightarrow{0} \mathbb{Z}_p \xrightarrow{0} \mathbb{Z}_p \xrightarrow{0} \mathbb{Z}_p \xrightarrow{0} \mathbb{Z}_p \xrightarrow{0} \dots$$

The homology of this complex is the cohomology of $S^{\infty}/\mathbb{Z}_p$ with $\mathbb{Z}_p$ coefficients, so we see that $H^k(S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p) \cong \mathbb{Z}_p$ for all $k \geq 0$.

Note that $p$ is prime (not stated in the post, but it is in the book), so $\mathbb{Z}_p$ is a field. As such, the Künneth theorem tells us that

$$H^k(S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p) \cong \bigoplus_{i+j = k}H^i(S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p)\otimes_{\mathbb{Z}_p}H^j(S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p).$$

As $\mathbb{Z}_p\otimes_{\mathbb{Z}_p}\mathbb{Z}_p \cong \mathbb{Z}_p$, we see that

$$H^k(S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p) \cong \mathbb{Z}_p^{k+1}.$$

In particular, $H^k(S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p) \neq 0, \mathbb{Z}_p$ for $k > 0$.

---

Note, the Künneth theorem actually gives an isomorphism of rings

$$H^*(S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p) \cong H^*(S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p)\otimes_{\mathbb{Z}_p}H^*(S^{\infty}/\mathbb{Z}_p; \mathbb{Z}_p).$$

Answer 2

One definition of group cohomology is that $H^n(G,M)$ (for the ab. group $M$, with trivial $G$-action) is that it is the cohomology of $K(G,1)$ with coefficients in the ab. group $M$.

If you want to regard this as a theorem instead, you have to begin from some other definition of group cohomology. Which one are you starting with?

If $G = G_1\times G_2$, then $K(G,1) = K(G_1,1)\times K(G_2,1)$, and so you can apply Kunneth to compute the cohomology (in terms of the cohomology of $G_1$ and $G_2$ separately).

Added: By the way, I noticed that in the related column on the right, the first link is Why is the cohomology of a $K(G,1)$ group cohomology?. Have you looked at the answers there?