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I'm seeking a normal subgroup $H$ of a group $G$ such that for some element $a \in G$, we have that $aHa^{-1} \subset H$ yet not $aHa^{-1} = H$.

Yet it seems to me this is impossible since $|aHa^{-1}| = |H|$ since if $f: H \rightarrow aHa^{-1}$ s.t. $f(h_i) = ah_ia^{-1}$, then $f$ is surjective (any $ah_ia^{-1} \in aHa^{-1}$ gets mapped to by $f(h_i)$) and $f$ is injective (if $h_i \ne h_j$ then $ah_ia^{-1} \ne ah_ja^{-1}$ since to assume otherwise would imply $a^{-1}(ah_ia^{-1})a = a^{-1}(ah_ja^{-1})a$ so that $h_i = h_j$ absurdly).

How -- given that such a bijection $f$ exists between $H$ and $aHa^{-1}$ -- could it be that there exists a normal subgroup $H$ of $G$ s.t. $aHa^{-1}$ is a proper subset of $H$?

EDIT: In fact, for any normal $H$ and for any $a \in G$, we have that $aHa^{-1} = Haa^{-1} = H$ so that it seems fundamentally impossible for this reason alone that $aHa^{-1} \subset H$. Here it doesn't seem to matter if $H$ is finite or infinite, doesn't it?

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    You answered your own question. It is not possible!2012-10-20
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    @Ittay: $|A|=|B|$ and $A\subseteq B$ does not imply that $A=B$ if $A$ is infinite.2012-10-20
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    It is not possible to find a normal subgroup $H$ of $G$ with that property. But there exist such examples in which $H$ is not normal in $G$.2012-10-20

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