Trying to show that : Show that a finite abelian group $M$ admits a basis as a $\Bbb Z$ -module if and only if $M =$ {$0$}.
The reverse direction being trivial, I am not sure what the forward direction means as I am new to the concept of a module. Any tips or clarifications much appreciated.
If $M$ is an $R$-module, a collection $(m_i)_{i \in I}$ of elements of $M$ forms a basis for $M$ (as a $R$-module) if and only if every element $m \in M$ can be represented uniquely as $m = \sum_{i \in I} r_i m_i$ for some $r_i \in R$ (where all but finitely many $r_i$ are required to vanish).
If $M$ is a $\mathbb{Z}$-module which admits a basis $(m_i)_{i \in I}$ then in particular if we choose some $i_0 \in I$ and look at the elements $(k \cdot m_{i_0})_{k \in \mathbb{Z}}$, they must all be distinct and so $M$ must be infinite.