Question: Group $G$ is abelian and finite. $\langle g\rangle = G$. $p$ is order of $G$ (and $\langle g\rangle$). $p=mn$, $m > 1$, $n > 1$. Why $\langle g^m\rangle < G$ (not $\langle g^m\rangle \le G$)?

Answer: 
Basic result in theory of cyclic groups: $\langle g\rangle =G\implies |g^m|=\dfrac n{\operatorname {gcd}(n,m)}$, where $n=|G|$.
As $|\langle g^m \rangle | = \dfrac {|G|} m$ and $m > 1$, so $|\langle g^m \rangle | < |G|$.