Why does $(ghg^{-1})^{-1}=(g^{-1})^{-1}h^{-1}g^{-1}$?

Question

I have read countless proofs that $gHg^{-1}$ is a subgroup of H but there is one step in the part about proving the inverse which I just cannot get my head around:

$(ghg^{-1})^{-1}=(g^{-1})^{-1}h^{-1}g^{-1}$

I just can't understand why it isn't:

$(ghg^{-1})^{-1}=g^{-1}h^{-1}(g^{-1})^{-1}$

Obviously there is something very fundamental/simple I don't understand so would appreciate if someone could explain this step to me.

Answer 1

Well, let's multiply them!

\begin{align} & \ \ \ \ \ (ghg^{-1})\cdot(g^{-1})^{-1}h^{-1}g^{-1}\\ &= gh(g^{-1}(g^{-1})^{-1})h^{-1}g^{-1}\\ &= gh(1)h^{-1}g^{-1}\\ &= g(hh^{-1})g^{-1}\\ &= g(1)g^{-1}\\ &= gg^{-1}\\ &=1 \end{align}

So really, the inverse of $ghg^{-1}$ is $(ghg^{-1})^{-1}=(g^{-1})^{-1}h^{-1}g^{-1}$.

Try multiplying $ghg^{-1}$ by what you thought the inverse was, $g^{-1}h^{-1}(g^{-1})^{-1}$, and you'll see the terms won't cancel, and the expression is not (necessarily) $1$.

Answer 2

You should know the fact that

For $a,b\in G$, $$(ab)^{-1}=b^{-1}a^{-1}$$

Otherwise, try to prove it.
So, $(ghg^{-1})^{-1}=((gh)g^{-1})^{-1}=(g^{-1})^{-1}(gh)^{-1}=(g^{-1})^{-1}h^{-1}g^{-1}$.