TITLE: An Example where $gHg^{-1} \ne g^{-1}Hg$
QUESTION [6 upvotes]: I was trying to think of an example where $gHg^{-1} \ne g^{-1}Hg$.  I couldn't think of one, but I am curious if the following reasoning demonstrates that, at the very least, such an example must exist:
Let $H$ be a non-normal subgroup of $G$ and let $g \in G - H$.  Then it is possible that $gHg^{-1} \ne g^{-1}Hg$ since if we supposed otherwise we would have
$$\begin{align}
gHg^{-1}   & = g^{-1}Hg \\
g^2Hg^{-1} & = Hg\\
g^2        & = e\\
g          & = e
\end{align}$$
which is impossible since we assumed $g \in G - H$.

REPLY [2 votes]: Or you could argue along the following lines, which is not so far from what you started to do.
If $gHg^{-1} = g^{-1}Hg$, then (multiplying both sides from the left by $g$ and from the right by $g^{-1}$), $g^{2}Hg^{-2} = H.$ This implies that $g^{2}$ is in the normalizer of $H$ 
(I hope you have done the normalizer in your course). So we need to find an example of a group $G$ with a subgroup $H$ and an element $g \in G$ such that $g^{2}$ does not normalize $H.$ Well, if $G$ has odd order, then $g$ is always  power of $g^{2}$ (we have $g^{2n+1} = e$ for some $n$), so if $g^{2}$ normalizes $H$, $g$ will too.Hence it is enough to find a group $G$ of odd order with a subgroup $H$ which is not normal. There is a non-Abelian group $G$ of order $21$ which has a Sylow $3$-subgroup $H$ which is not normal. We can take $g$ to be any element of order $7$ in $G$, and we will have $gHg^{-1} \neq g^{-1}Hg.$