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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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HINT: Let $G$ be the multiplicative group of non-singular $2\times 2$ real matrices. Let $H$ be the subgroup of $G$ generated by the matrix $\pmatrix{1&1\\0&1}$. Find a diagonal matrix $g\in G$ such that $gHg^{-1}$ is a proper subgroup of $H$.

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Hint: you can find an example with $H = \mathbb{Z}$ and $aHa^{-1} = 2\mathbb{Z} \subset \mathbb{Z}$.

It's true that $f$ is a bijection, but $f$ has nothing to do with the inclusion $aHa^{-1} \subset H$. This inclusion is injective but there's no reason for it to be surjective if $H$ is infinite (as in the above example).

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    More explicitly, $f$ is usually $g \mapsto a^{-1}ga$.2012-10-20
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    @Lord: why? That doesn't exhibit conjugation as a left action.2012-10-20
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    Disregard that, it must have been an ignorant monkey creeping under my skin.2012-10-20
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    Qiaochu -- since all of the supersets of $\mathbb{Z}$ that I can think of are abelian (i.e., $\mathbb{R}$, $\mathbb{C}$), I can't come up with an $a \in G$ s.t. $aHa^{-1} = 2\mathbb{Z}$ since under such situations $aHa^{-1} = aa^{-1}H = H = \mathbb{Z}$.2012-10-20
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    @George: The group $H$ in my answer isn’t $\Bbb Z$, but it is isomorphic to $\Bbb Z$.2012-10-20
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    @George: $\mathbb{Z}$ is a subgroup of every group with an element of infinite order. That's a lot of groups! Think about matrix groups, for example (which are usually nonabelian).2012-10-20
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Any normal subgroup necessarily satisfies $aHa^{-1} = Haa^{-1} = H$. However, your objection on the basis of cardinality is only compelling for finite subgroups.

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    In fact, for any normal H and for any a∈G, we have that aHa−1=Haa−1=H so that it seems fundamentally impossible for this reason alone that aHa−1⊂H. Here it doesn't seem to matter if H is finite or infinite, doesn't it?2012-10-20
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    @George Correct. The question is much more interesting if you drop the restriction that $H$ is normal, then the other answers apply.2012-10-20