Group Theory Inverse Proof

Question

I'm given the proposition: Let $G$ be a group with elements $g$ and $h$. If $g*h=e$, then $h*g=e$, where $e$ is the identity element.

I'm not entirely sure how to prove this one, though, so some help would be nice.

The definition of an inverse in group theory that I was given was "There exists an element h such that $g*h=h*g=e$, where $h=g^{-1}$," but I don't think my professor is looking for that obvious solution.

Answer 1

$h$ does not necessarily the inverse of $g$. If $g*h=e$, then $$h*g=h*(e*g)=h*(g*h)*g=(h*g)*(h*g)=(h*g)^2$$ and, since $G$ is a group, applying the inverse of $h*g$ on the equation, $h*g=e$.

Answer 2

Hint: Multiply the equation $gh = e$ by $g^{-1}$ on the left and $g$ on the right.

Answer 3

Take it that $gh = e$. Therefore $g^{-1}gh = g^{-1}$, and hence $eh = g^{-1}$ and $e = g^{-1}h^{-1}$. Then all you need is $e^{-1} = e$ and $(hg)^{-1} = g^{-1}h^{-1}$ to finish the proof. These are elementary claims.

Answer 4

$$hg = hge = hghh^{-1} = heh^{-1} = hh^{-1} = e.$$