Suppose that $M$ is a $\Bbb Z/36 \Bbb Z$-module such that $\bar 6 \cdot M = \{0\}$. Prove that $M$ is Noetherian $\iff$ $M$ is Artinian.
I managed to prove the forward direction: if $M$ is Noetherian, then it's finitely generated, and as $\Bbb Z/36 \Bbb Z$ is Artinian, so is $M$.
I must be missing something obvious. Can I get a hint?
Since $\overline 6\in\operatorname{Ann}(M)$, there is an induced action of the ring $(\mathbb Z/36\mathbb Z)/(\overline 6) = (\mathbb Z/36\mathbb Z)/(6\mathbb Z/36\mathbb Z)\cong \mathbb Z/6\mathbb Z$ on $M$, i.e. $M$ may be considered as a $\mathbb Z/6\mathbb Z$-module, and under this action, $M$ has the same submodules as under the original action, so in particular $M$ is Noetherian (resp. Artinian) under the $Z/36\mathbb Z$ action if and only if it is Noetherian (Artinian) under the $Z/6\mathbb Z$ action.
Now, by the Chinese Remainder Theorem, $\mathbb Z/6\mathbb Z \cong \mathbb Z/2\mathbb Z \oplus \mathbb Z/3\mathbb Z$, so we may consider $M$ as a $(\mathbb Z/2\mathbb Z \oplus \mathbb Z/3\mathbb Z)$-module. Set $M_2=(1,0)M$ and $M_3=(0,1)M$. Show that $M=M_2 \oplus M_3$, that $M_2$ (resp. $M_3$) can be considered as a $\mathbb Z/2\mathbb Z$-vector space (resp. $\mathbb Z/3\mathbb Z$-vector space), and use the fact that for vector spaces the Artinian and Noetherian conditions are both equivalent to finite dimensionality.