Can someone please verify whether my proof is logically correct? :) Note that it is not assumed that $\left \langle a\right \rangle \cap \left \langle b\right \rangle=\left \{ e \right \}$.
If $a$ and $b$ are generators of a group $G$, then show that $\left \langle a\right \rangle \cap \left \langle b\right \rangle$ is a subgroup of both $\left \langle a\right \rangle$ and $\left \langle b\right \rangle$.
Proof:
Let $\left \langle a \right \rangle \cap \left \langle b \right \rangle =\left \{ c^{k}:c^{k}\in \left \langle a \right \rangle,\left \langle b \right \rangle \right \}$.
Since $e\in \left \langle a \right \rangle$ and $e\in \left \langle b \right \rangle$, then $e\in \left \langle a \right \rangle \cap \left \langle b \right \rangle$ by definition.
Let $c^{k},c^{m}\in \left \langle a \right \rangle \cap \left \langle b \right \rangle$. Then $c^{k},c^{m}\in \left \langle a \right \rangle $ and $c^{k},c^{m}\in \left \langle b \right \rangle $. Then $c^{k}(c^{m})^{-1}=c^{k}c^{-m}=c^{k-m}\in \left \langle a \right \rangle \cap \left \langle b \right \rangle$, since $c^{k}(c^{m})^{-1}=c^{k}c^{-m}=c^{k-m}\in \left \langle a \right \rangle, \left \langle b \right \rangle$. Therefore, $\left \langle a \right \rangle \cap \left \langle b \right \rangle \leq \left \langle a \right \rangle$ and $\left \langle a \right \rangle \cap \left \langle b \right \rangle \leq \left \langle b \right \rangle$. $\square$
If $H$ and $K$ are subgroups of $G$, then $H \cap K$ is a subgroup of both $H$ and $K$. The fact that $H$ and $K$ are generated by a single element is irrelevant.