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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 $g \in G - H$.

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First, $g^2 H g^{-1} = H g$ does not imply $g^2 = e$. Second $g^2 = e$ does not imply $g = e$. But +1 for providing your attempted answer. –  Michael Joyce Oct 5 '12 at 16:35
    
You can't have tried very hard to find an example, if you haven't checked the smallest non-abelian group. –  Chris Eagle Oct 5 '12 at 16:35
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@ChrisEagle Maybe he doesn't know what that is. George, it is $S_3$, the symmetric group on 3 elements, a group of order 6. –  Graphth Oct 5 '12 at 16:38
    
Sometimes one does prove existence indirectly. However, apart from issues of detail, there would still be the problem of showing that there is a pair $(G,H)$ with $H$ not normal. –  André Nicolas Oct 5 '12 at 16:40
    
$g^2 = e$ doesn't imply $g = e$ since in $S_3$ we have that $(1\,2)^2 = e$, for instance. Let $H = \{e, (1\,2)\} < S_3$ and consider that if $g = (1\,2\,3)$, then $gHg^{-1} \ne g^{-1}Hg$ –  George Oct 5 '12 at 16:49

3 Answers 3

You won't come up with a counterexample when $G$ is abelian (why?). So your best bet is to try it with $G$ being non-abelian, and the simplest non-abelian group is $S_3$.

Moreover, you are bound to fail if $g = g^{-1}$ (why?). So let's try letting $g = (1,2,3)$, in cycle notation. For our subgroup $H$, let's pick one which involves elements that don't commute with $g$. So we're going to want a transposition in our subgroup. So let's take $H = \{ \text{id}, (1,2) \}$, the subgroup consisting of the identity permutation and the transposition interchanging $1$ and $2$.

Try computing $g H g^{-1}$ and $g^{-1} H g$ for this example. (It will work; but if it hadn't, the best thing to do would be to try to understand why it didn't work and try to come up with the next simplest example that wouldn't be sure to fail based on the reasons you know to this point.)

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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 $gHg^{-1} \neq g^{-1}Hg.$

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Note that $gHg^{-1}\ne g^{-1}Hg$ is equivalent to $g^2 Hg^{-2}\ne H$. Consider $G=S_3$ and $H$ a cyclic subgroup of order 2. Conjugating $H$ with any $G$ of order 3 will produce a conjugate different from $H$. Note that every 3-cycle is the square of its inverse.

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