Suppose that G is a group of order 6 and that $a,b \in G,$ where $a \neq b.$ If both $a$ and $b$ have order 2, justify that G is generated by $a$ and $b$.
By the order requirement, $a^2=e$, and $b^2=e$ and 2 is the smallest number for which this is the case. Namely, $a \neq e,$ and $b \neq e.$ So far, our group consists of $G=\{e,a,b \}.$ It must be closed under multiplication, so $ab \in G$. Also, a quick computation shows that $(ab)^{-1}=ba,$ so $ba \in G.$ However, I need one more element, and I need to prove that all of these are distinct and exhaust the possibilities and I have no idea how to do that. Any help would be appreciated.
Thanks.
The subgroup $\langle a, b \rangle$ has order $>3$ since $\langle a \rangle$ is a proper subgroup of $\langle a, b \rangle$. By Lagrange's theorem, the order of $\langle a, b \rangle$ can only be $6$.