Do Eilenberg-MacLane spaces $K(G, n), n \ge 2$ always have infinite cohomological dimension?

Question

Ben Steffan recently asked whether a $4$-manifold can be an Eilenberg-MacLane space $K(G, n), n \ge 2$ (and $G$ an arbitrary abelian group). Personally I would guess that the answer is no, and moreover I would guess that this is not specific to $4$-manifolds: more precisely, I would guess that for $G$ nontrivial and $n \ge 2$, $K(G, n)$ has infinite cohomological dimension, so cannot be a manifold (or CW complex) of any (finite) dimension.

Question: Is this true?

Classical computations with spectral sequences imply that this is true if $G$ is cyclic, hence if it's finitely generated, and rational homotopy stuff implies that this is true if $G$ is a $\mathbb{Q}$-vector space, but I don't know what happens in general. Since the cohomology $H^m(K(G, n), A)$ can be identified with the set of cohomology operations $H^n(-, G) \to H^m(-, A)$, an equivalent question is whether for $n \ge 2$ there are always nonvanishing cohomology operations on a cohomology class of degree $n$ with coefficients in $G$, sending it to another cohomology class of arbitrarily high degree with some coefficients.

At least for $n$ even here is a potential candidate for such an operation: I believe there should be an "external cup power" operation $H^n(-, G) \to H^{nk}(-, G^{\otimes k})$ for any $k \ge 1$. Unfortunately $G^{\otimes k}$ can vanish for $k \ge 2$, for example if $G = \mathbb{Q}/\mathbb{Z}$.

What would be ideal is a version for $n \ge 1$ of the argument for $n = 1$ that if $H \le G$ is a subgroup then $\text{cd}(K(H, 1)) \le \text{cd}(K(G, 1))$, which would let us reduce to the cyclic case. I guess the natural idea is to look at the Serre spectral sequence associated to the fibration $K(G/H, n-1) \to K(H, n) \to K(G, n)$; I've never been fluent with spectral sequences but offhand this only seems to imply that if $K(H, n)$ has infinite cohomological dimension then either $K(G, n)$ or $K(G/H, n-1)$ must also have infinite cohomological dimension.

Answer 1

This is false.

I assume that you still only care about $n \geq 2$ since any $K(\mathbb{Z}, 1)$ is a counterexample. It is a standard application of the Serre spectral sequence and Serre class theory that $K(G, n)$ has infinite cohomological dimension if $G$ is finitely generated, as you say. But then you state that "rational homotopy stuff implies that this is true if $G$ is a $\mathbb{Q}$-vector space", and this is false: The computation of $H_*(K(\mathbb{Q}, n))$ is another standard application of the Serre spectral sequence, and it yields that $$ \tilde{H}_k(K(\mathbb{Q}, n)) \cong \begin{cases} \mathbb{Q} & n \text{ odd},\ k = n \\ \mathbb{Q} & n \text{ even},\ k > 0 \text{ and } n \mid k \\ 0 & \text{else} \end{cases} $$ By the UCT, we then find that $$ \tilde{H}^k(K(\mathbb{Q}, 2n + 1)) \cong \begin{cases} \mathbb{Q} & k = 2n + 1 \\ \operatorname{Ext}(\mathbb{Q}, \mathbb{Z}) & k = 2n + 2 \\ 0 & \text{else} \end{cases} $$ so $K(\mathbb{Q}, 2n + 1)$ is of finite cohomological dimension.

As for cohomology operations, you need to be more careful: If $G$ is the additive group of a ring, then sure, you'll have cup powers. These will take the form $H^n({{-}}; G) \Rightarrow H^{kn}({{-}}; G)$, and will give rise to elements in $H^{kn}(K(G, n); G)$. But they can be zero: For instance this must be the case if $G$ has no 2-torsion and $n$ is odd since the cup product is graded-commutative (I'll leave it as an exercise to ponder the connection to the cohomology of $K(\mathbb{Q}, n)$ stated above). If $G$ is not the additive group of a ring, e.g. when $G$ is a Prüfer group, then there is no reason to expect something like cup powers to exist at all. Producing other (provably non-zero) cohomology operations is generally much more difficult.

The natural generalization to ask is thus "For which combinations of $G$ and $n$ is $K(G, n)$ of finite cohomological dimension?" I do not know the answer to this question, but this fantastic answer resolves the weaker question of "For which combinations of $G$ and $n$ is $K(G, n)$ of homological dimension $n$:" In this case it turns out that $(\mathbb{Q}, 2n + 1)$ and $(P, 2n)$ where $P$ is a product of Prüfer $p$-groups for odd primes $p$ are the only positive answers.