Originally Posted by
Black I think this will work.
Let $\displaystyle a,b \in G$.
If $\displaystyle a=b,$ then $\displaystyle ab=ba$.
If $\displaystyle a=b^{-1},$ then $\displaystyle ab=ba=e$.
So suppose that the two aren't equal and inverses of each other. Then $\displaystyle ab$ is an element that is not equal to $\displaystyle e,a,$ or $\displaystyle b$. This goes for $\displaystyle ba$ as well.
Now suppose that $\displaystyle ab \not= ba$. Then we have
$\displaystyle G=\{e,a,b,ab,ba \},$
which means that $\displaystyle a^{-1},b^{-1} \in \{ab,ba\}.$ If $\displaystyle a^{-1}=ab$ and $\displaystyle b^{-1}=ba,$ then
$\displaystyle a^2b=e \Longrightarrow a^2=ba \Longrightarrow a=b,$
which is a contradiction. The case for $\displaystyle a^{-1}=ba, \, b^{-1}=ab$ is similar.