Let $M$ be $\mathbb{Z}/16\mathbb{Z}\oplus \mathbb{Z}/25\mathbb{Z}$ as a $\mathbb{Z}$-module. I want to compute $\text{soc}(M)$, where $\text{soc}(M)$ is the sum of simple $\mathbb{Z}$-modules of $M$.
All $\mathbb{Z}$-submodules of $\mathbb{Z}/16\mathbb{Z}$ are $0, \mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}, \mathbb{Z}/8\mathbb{Z}, \mathbb{Z}/16\mathbb{Z}$, and the unique simple $\mathbb{Z}$-submodule of $\mathbb{Z}/16\mathbb{Z}$ is $\mathbb{Z}/2\mathbb{Z}$. Similarly, the unique simple $\mathbb{Z}$-submodule of $\mathbb{Z}/25\mathbb{Z}$ is $\mathbb{Z}/5\mathbb{Z}$. So $\text{soc}(M)$ is $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}$.
Is this correct?
Thanks for your help.
As user Omnomnomnom points out, when making the claim that ''all $\mathbb{Z}$-modules of $\mathbb{Z}_{16}$ are etc'', what you proceed to enumerate must really be a list of actual submodules, not merely of objects isomorphic to submodules; in this sense, $\mathbb{Z}_2$ for instance is not itself a submodule of $\mathbb{Z}_{16}$, but merely isomorphic to the submodule $8 \mathbb{Z}_{16}=8\mathbb{Z}/16\mathbb{Z}$, so let us be rigorous with our language.
Your reasoning needs one more added detail to ensure that all the simple submodules actually arise from one or the other of the two summands (and that there is no possibility of a simple submodule situated ''obliquely'' so to speak). I will only be presenting the relevant propositions without any proofs so do write back if you would like more details.
First, a few notions and conventions: fixing an arbitrary set $\Omega$, we say $(G, \cdot, \cdot)$ is an $\Omega$-group if the first multiplicative law $\cdot: G \times G \to G$ is an internal one rendering the structure $(G, \cdot)$ into a group and respectively the second one $\cdot: \Omega \times G \to G$ is an external action of $\Omega$ on $G$ via group endomorphisms, in other words such that ${}^{\lambda}(xy)={}^{\lambda}x{}^{\lambda}y$ for any $\lambda \in \Omega, x, y \in G$ (where the result of the action of the operator $\lambda$ on $x$ is denoted by ${}^{\lambda}x$). One can go on to very naturally define morphisms of $\Omega$-groups and to ultimately set up the category of $\Omega$-groups. Although somewhat abstract, they do serve as a generalization of modules to the non-commutative setting (and are actually categorically equivalent to the category of modules over a particular ring in the commutative case).
(Normal) $\Omega$-subgroups of $\Omega$-groups are defined as in the usual case of groups (with no external operators), and we introduce the notion of subquotient of $\Omega$-group $G$ to mean any $\Omega$-group isomorphic to a quotient $H/K$, where $H \leqslant_{\Omega} G$ is a $\Omega$-subgroup of $G$ and $K \trianglelefteq_{\Omega} H$ is a normal $\Omega$-subgroup of $H$.
Now on to the announced
This easily applies to your case since $(16, 25)=1$ and the order of any subquotient common to both $\mathbb{Z}_{16}$ and $\mathbb{Z}_{25}$ will be a common divisor of the two respective orders.
To finalize, as both commentators above have properly pointed out, the correct way to express the result is:
$$\mathrm{Soc}(\mathbb{Z}_{p^m} \times \mathbb{Z}_{q^n})=p^{m-1} \mathbb{Z}/p^m \mathbb{Z} \times q^{n-1} \mathbb{Z}/q^n \mathbb{Z}$$
where $p, q$ are two distinct primes and $m, n \in \mathbb{N}^*$ are nonzero natural exponents.
The proposition presented above can be generalized to a finite family of $\Omega$-groups such that pairwise they only have the trivial group as a common subquotient, and this allows one to obtain the following general result for arbitrary $m \in \mathbb{N}^*$
$$\mathrm{Soc}\ (\mathbb{Z}_m) \approx \prod_{p \in \Pi(m)} \mathbb{Z}_p$$
where we have used the conventions of notation
$$\mathbb{P}=\{p \in \mathbb{N}^*|\ p\ \text{prime}\}$$
to denote the set of all (nonzero natural) primes and $\Pi(n)=\{p \in \mathbb{P}|\ p|n\}$ to denote the set of all prime divisors of arbitrary $n \in \mathbb{Z}$.