Prove that $ a^6 = b^{11} $

Question

Let $a, b \in G$, where $G$ is a group, and $|a| = 12$ and $|b|=22$. If $\langle a \rangle \cap \langle b \rangle \neq \{e\}$, prove that $a^6 = b^{11}$.

Answer 1

Notice that the order of $\langle a \rangle \cap \langle b \rangle$ divides both $|a|$ and $|b|$ because it's a subgroup of both $\langle a \rangle$ and $\langle b \rangle$.

Answer 2

Hint:

If $x\in\langle a\rangle$ and $y\in\langle b\rangle$ with $x=y$ then $x$ and $y$ must have equal order.

Answer 3

The order of $\langle a\rangle \cap \langle b\rangle$ has to divide both $12$ and $22$. If this subgroup is not trivial, then it must have order $2$-which elements of $\langle a\rangle$ and $\langle b\rangle$ have order $2$?