TITLE: Abelian group zeta function
QUESTION [6 upvotes]: Let $s \in \mathbb{C}$. What's known about
$$\zeta_{\mathrm{ab}}(s) := \sum_G \frac{1}{o(G)^s} \tag{1}$$
where the sum is over all finite abelian groups $G$ up to isomorphism?
By the primary decomposition theorem for finite abelian groups, I feel like $\zeta_{\mathrm{ab}}(s)$ should have an Euler product
$$\zeta_{\mathrm{ab}}(s) = \prod_C \Big(1 - \frac{1}{o(C)^s}\Big)^{-1} \tag{2}$$
over all (nontrivial) cyclic groups of prime-power order.
I would be curious to know, e.g., the abscissae of convergence of (1)
and (2), whether $\zeta_{\mathrm{ab}}(s)$ has a meromorphic extension
to $\mathbb{C}$, what's known about its zeroes and poles, etc.

Some observations:
$$\zeta_{\mathrm{ab}}(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{2}{4^s} + \frac{1}{5^s} + \frac{1}{6^s} + \frac{1}{7^s} + \frac{3}{8^s} + \frac{2}{9^s} + \ldots = \sum_{n=1}^\infty \frac{\nu_{\mathrm{ab}}(n)}{n^s} \tag{3}$$
where $\nu_{\mathrm{ab}}(n)$ is the number of abelian groups of order $n$ (OEIS:688). Since $\nu_{\mathrm{ab}}(n)$ is multiplicative, I suppose that makes (3) the Dirichlet series for $\zeta_{\mathrm{ab}}(s)$.
In terms of the partition function, we have
$$\nu_{\mathrm{ab}}(n) = \prod_{p \mid n} p(\operatorname{ord}_p(n)).$$
In particular $\nu_{\mathrm{ab}}(p^k) = p(k)$ (the notation is unfortunate).
Formally, it follows that
$$\zeta_{\mathrm{ab}}(s) = \prod_p \sum_{k \ge 0} p(k) p^{-ks} \tag{4}$$
wherever it converges. Any insights or references would be appreciated!

REPLY [4 votes]: I'll write $a_n$ for the number of Abelian groups of order $n$ up to isomorphism.
Then indeed $a_n$ us multiplicative: $a_{mn}=a_ma_n$ for $\gcd(m,n)=1$, so the zeta
function has an Euler product.
For a prime $p$, the groups of order $p^k$ have the form $C_{p^{t_1}}\times\cdots\times
C_{p^{t_r}}$ uniquely for $t_1\ge\cdots\ge t_r\ge1$ and $t_1+\cdots+t_r=k$.
They correspond to the partitions of $k$, so $a_{p^k}=p(k)$, the $k$-th partition
number (alas I am using $p$ for two different things here).
So
$$\sum_{k=0}^\infty\frac{a_{p^k}}{p^{ks}}
=\sum_{k=0}^\infty\frac{p(k)}{p^{ks}}=\prod_{j=1}^\infty\left(1-\frac{1}{p^{js}}\right)^{-1}$$
by a well-known identity. We conclude that
$$\zeta_{\text ab}(s)=\prod_p\prod_{j=1}^\infty\left(1-\frac{1}{p^{js}}\right)^{-1}
=\prod_{j=1}^\infty\prod_p\left(1-\frac{1}{p^{js}}\right)^{-1}
=\prod_{j=1}^\infty\zeta(js)$$
at least formally.