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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 $g \in G - H$.

  • 2
    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.2012-10-05
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    You can't have tried very hard to find an example, if you haven't checked the smallest non-abelian group.2012-10-05
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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 6.2012-10-05
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    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.2012-10-05
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    $g^2 = e$ 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$2012-10-05
  • 0
    A subgroup is called *Malnormal* if $H^g\cap H\neq 1\Rightarrow g\in H$. Malnormal subgroups are important in geometric group theory. Now, if you find a group with a malnormal subgroup containing no elements $g\in G$ with $g^2\in H$ then you are done. If you know what a free group is, then any cyclic subgroup of a free group which is not a proper subgroup of another cyclic subgroup is malnormal, and this works.2012-10-08

3 Answers 3