This is a clarification question about Exercise 8.1 in section VIII of Silverman's Arithmetic of Elliptic curves. The original exercise is:
Let $E/K$ be an elliptic curve, let $m \geq 2$ be an integer, let $\mathcal H_K$ be the ideal class group of $K$ and let $$S = \{ \nu \in M_K^0 : E \text{ has bad reduction at } \nu \} \cup \{ \nu \in M_K^0 : \nu(m) \neq 0 \} \cup M_K^\infty.$$ Assume that $E[m] \subset E(K)$. Prove the following quantitative version of the weak Mordell-Weil theorem: $$rank_{\mathbb{Z}/m \mathbb{Z}} E(K)/mE(K) \leq 2 \# S + 2 rank_{\mathbb{Z}/m\mathbb{Z}} \mathcal H_K[m].$$
I can prove this statement if $m$ is a prime. My question is, what does $rank_{\mathbb{Z}/m\mathbb{Z}}$ even mean if $m$ is not prime? For instance, what is the $\mathbb{Z}/6\mathbb{Z}$-rank of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/ 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z} \simeq \mathbb{Z}/ 2 \mathbb{Z} \times \mathbb{Z} / 6 \mathbb{Z}$? This group contains "slightly more" than one copy of $\mathbb{Z}/6\mathbb{Z}$, but not quite as much as two copies.
Someone suggested to me that for a finite group $G$ that is also a $\mathbb Z/m \mathbb Z$ module $r = rank_{\mathbb Z/ m \mathbb Z}(G)$ could be the number of generators needed to generate $G$ over $\mathbb Z/ m \mathbb Z$. You can convince yourself such a group $G$ can be written as a product of $r$ cyclic groups of order dividing $m$, so this is sort of similar to the free case where we'd have $G \simeq (\mathbb Z/ m \mathbb Z)^r$. (In my above example $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 6 \mathbb Z$ would have rank $2$.)
I think this definition makes the statement in Silverman's question correct. To prove it I needed that $H \subset G$ implies $rank_{\mathbb Z/ m \mathbb Z}(H) \leq rank_{\mathbb Z/ m \mathbb Z}(G)$ (when both are $\mathbb Z/ m \mathbb Z$-modules). Indeed if $H \subset G$ then $H$ is a "smaller" product of cyclic groups of order dividing $m$ i.e. $H$ can be written as $\leq rank_{\mathbb Z/ m \mathbb Z}(G)$ cyclic groups, each of order less than a corresponding cyclic group of $G$. The ingredients I used to show this are (1) the fundamental theorem of finitely generated abelian groups (2) using multiplication by $p^n$ morphisms for $p \mid m$ to isolate the p-primary components of $G$ and $H$ (3) reducing the $p$-primary groups mod $p$ to make the claim $rank(H) \leq rank(G)$ obvious.
Then the proof proceeds as in Prop. VIII.1.6 of Silverman (see also this lecture for more details).