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$H<G,aHa^{-1}<G$. Then, $H$ is isomorphic to $aHa^{-1}$.

I want to show $aha^{-1}\in H,\forall h\in H,a\in G$, but I cannot figure it out. Any hint is appreciated.

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Being isomorph isn't the same as being equal. You have to construct an Isomorphism from $H$ to $aHa^{-1}$. – Stefan Feb 19 '13 at 8:32
Thank you very much, @Stefan. I was confused with two cases. – Sam Feb 19 '13 at 8:38
$h\mapsto aha^{-1}$ is an isomorphism $H\to aHa^{-1}$ and this fact can be expressed as: The group $G$ operates on the set of its subgroups by conjugation. The normal subgroups are precisely the fixed points under this operation. – Hagen von Eitzen Feb 19 '13 at 8:56
Thanks a lot, @Hagen, for your further explanation. – Sam Feb 19 '13 at 9:07
So, Sam, can you see now how to prove the isomorphism? – Gerry Myerson Feb 19 '13 at 10:04
up vote 2 down vote accepted

To show that $H\cong gHg^{-1}$, you need to show that for any $a\in G$, the conjugation $$\kappa_a : G\to G,\quad g\mapsto aga^{-1}$$ is an automorphism of $G$. The proof is easy: For $g,h\in G$ we have $$\kappa_a(gh) = agha^{-1} = ag(a^{-1}a)ha^{-1} = (aga^{-1})(aha^{-1}) = \kappa_a(g)\kappa_a(h)\text{,}$$ so $\kappa_a$ is a homomorpisms. To show that $\kappa_a$ is one-to-one, check $\kappa_a\circ\kappa_{a^{-1}} = \operatorname{id}$.

Now every subgroup $H$ is isomorphic to $\kappa_a(H) = aHa^{-1}$.

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Thank you very much, @azimut. – Sam Feb 20 '13 at 2:22

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