Structure theorem for profinite abelian groups

Question

If $G$ is a finitely generated profinite abelian group then $$ G\cong \prod_p \mathbb Z_p^{a(p)}\times\prod_n(\mathbb Z/n\mathbb Z)^{b(n)} $$ where $a(p), b(n)$ are integers and $a(p)=b(n)=0$ for all large enough $n$ and $p$ (see Ribes "Profinite groups" Theorem 4.3.5). If $G$ is a torsion-free profinite abelian group then $$ G\cong \prod_p\mathbb Z_p^{c(p)} $$ where the $c(p)$ are arbitrary cardinal numbers (see Ribes Theorem 4.3.3). If $G$ is a torsion profinite abelian group then $$ G\cong \prod_n (\mathbb Z/n\mathbb Z)^{c(n)} $$ where the $c(n)$ are arbitrary cardinal numbers with $c(n)=0$ for all large enough $n$ (see Ribes Corollary 4.3.9).

My question: Is every profinite abelian group of the form $$ \prod_p \mathbb Z_p^{a(p)}\times \prod_n (\mathbb Z/n\mathbb Z)^{b(p)} $$ where the $a(p)$ and $b(n)$ are arbitrary cardinal numbers? If the answer is positive, I'd like to see a proof (a reference suffices). If the answer is negative, what is a counter-example (with proof)?

Some comments: We know that a profinite abelian group $G$ is an inverse limit of finite abelian groups. Say $\{ G_i , \varphi_{ij}, I\}$ is a surjective inverse system of finite abelian groups. Then $G\cong \varprojlim_{i\in I} G_i$ is a profinite abelian group, and every profinite abelian group is of this form. Since the $G_i$ are finite and abelian, one can use the structure theorem for finite abelian groups on the $G_i$ and hope it "propagates" to the limit.

Answer 1

No. For me it's easier to think about the Pontryagin dual question: recall that Pontryagin duality $A \mapsto \hat{A} \cong \text{Hom}(A, S^1)$ restricts to a contravariant equivalence of categories between profinite abelian groups and torsion abelian groups (abstractly, these are the pro-category of finite abelian groups and the ind-category of finite abelian groups respectively, and the category of finite abelian groups is self-dual). We could also use $\mathbb{Q}/\mathbb{Z}$ here as the dualizing object. So, the Pontryagin dual question is whether every torsion abelian group is isomorphic to a direct sum

$$\bigoplus_p (\mathbb{Q}_p/\mathbb{Z}_p)^{\oplus a(p)} \oplus \bigoplus_n (\mathbb{Z}/n)^{\oplus b(n)}$$

of Prüfer $p$-groups and finite cyclic groups.

Every torsion abelian group $A$ is a direct sum $A \cong \bigoplus_p A_p$ of its $p$-primary parts, so we can address this question by working one prime at a time. Fixing a prime $p$ and restricting our attention to abelian $p$-groups WLOG, the question is whether every abelian $p$-group $A$ is a direct sum of copies of the Prüfer $p$-group $\mathbb{Q}_p/\mathbb{Z}_p$ and the finite cyclic groups $\mathbb{Z}/p^k $.

The theory of Ulm invariants implies that this is very far from true. Namely, as in this previous question, consider the abelian $p$-group with presentation

$$A = \langle x_0, x_1, x_2, \dots \mid px_0 = 0, p^k x_k = x_0 \rangle.$$

As Derek Holt says in a comment, quotienting by $x_0$ produces the direct sum $\bigoplus_k \mathbb{Z}/p^k$, so $A$ is a nontrivial extension

$$0 \to \mathbb{Z}/p \to A \to \bigoplus_k \mathbb{Z}/p^k \to 0.$$

For an abelian $p$-group $A$ its (first) Ulm subgroup is $U(A) = \cap_n p^n A$, the subgroup of divisible elements of $A$. For a direct sum of copies of the Prüfer $p$-group and finite cyclic groups the Ulm subgroup is given by the Prüfer $p$-group part. But the Ulm subgroup of the group $A$ above is (if I'm not mistaken) $U(A) = \langle x_0 \rangle \cong \mathbb{Z}/p$, so it cannot be isomorphic to a direct sum of the above form.

I can't say I particularly understand the profinite abelian group given by the Pontryagin dual $\hat{A}$, but dualizing the short exact sequence above, we can at least say that it's a nontrivial extension

$$0 \to \prod_k \mathbb{Z}/p^k \to \hat{A} \to \mathbb{Z}/p \to 0.$$

For more discussion of Ulm invariants I found Arturo Magidin's answer in this old thread helpful.