Let $G$ be a finite abelian group. Prove that for all $a,b \in G$, the set $$H = \langle a,b\rangle =\{a^m b^n | m,n \in \mathbb{Z} \}$$ is a subgroup of $G$.
I tried to check that the following $3$ criteria hold, but I couldn't proceed using the definition of $H$.
\begin{align} & e \in H \\ & ab \in H \\ & a^{-1} \in H \\ \end{align}
I prove the more general case for which your question is a special case.
By definition $\langle S\rangle:=\left\{\prod_{i=1}^ls_i:l\in\mathbb{N_0},\forall i\in\{1,\ldots,l\}\ ( s_i\in S \vee s_i^{-1}\in S)\right\}$, with the empty product equal to $e\in G$. Now clearly $\langle S\rangle\neq\varnothing$, since $e\in\langle S\rangle$ (even when $S=\varnothing$). Now suppose $a,b\in\langle S\rangle$. Thus for certain $l_1,l_2\in\mathbb{N_0}$, and $s_1,\ldots,s_{l_1},s'_1,\ldots,s'_{l_2}\in S$, $a=\prod_{i=1}^{l_1}s_i$ and $b=\prod_{i=1}^{l_2}s'_i$, thus $ab^{-1}=\prod_{i=1}^{l_1}s_i\left(\prod_{j=1}^{l_2}s'_j\right)^{-1}=\prod_{i=1}^{l_1}s_i\prod_{j=1}^{l_2}(s'_{l_2-j+1})^{-1}$. By looking once again at the definition of $\langle S\rangle$ we see $ab^{-1}\in\langle S\rangle$, as desired. Thus $\langle S\rangle\subset G$ is indeed a subgroup.