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Is there a group $G$, a subgroup $H$ and an element $x$, such that $xHx^{-1} \subset H$, but $xHx^{-1} \neq H$?

Thanks in advance.

Martin Sleziak
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D. N.
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    If they are finite then no as they have the same number of elements. – James Aug 08 '13 at 04:10
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    maybe consider $$ \langle x,a_0,a_1,...|\ a_{i+1}=xa_ix^{-1}\rangle $$ (i dont know) – yoyo Aug 08 '13 at 04:47
  • I like this question a lot. I hadn't seen it before, and I spent a good time thinking about it. – davidlowryduda Aug 08 '13 at 04:58
  • @yoyo Your presentation works. It is simply an HNN-extension of $F_{\infty}$ (and is actually isomorphic to $F_{\infty}$ by discarding $a_0$ and using $x$ along with $b_i:=xa_ix$, $i>0$, as your generating set). A simpler presentation would be $\langle a, t; t^{-1}at=a^2\rangle$, and again this is an HNN-extension. See my answer here for more details. – user1729 Aug 08 '13 at 18:10

1 Answers1

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Yes, there is such a group, subgroup, and element.

Consider the symmetric group of the integers $\text{Sym}(\mathbb{Z})$. You can think of this as the group of permutations on the integers. Let $H$ be the set of permutations that fix the negative integers. Let $x$ be the permutation satisfying $x(a) = a-1$.

Let's consider $x^{-1}Hx$. First $x$ shifts all the integers down, any element in $H$ fixes the integers in all negative slots, and $x^{-1}$ shifts the integers back up. This means that the integers in slots less than $1$ are all fixed.

Since the group of permutations that fix integers less than $1$ is a subgroup of the permutations that fix negative integers, we have that $x^{-1}Hx \subset H$. But $x^{-1}Hx$ doesn't contain any permutation that shifts the $0$'th entry, so $x^{-1}Hx \neq H$.

Note that I pivotally required an infinite nonabelian group. Any abelian group only has normal subgroups. If the group is finite, then $x^{-1}Hx$ will have the same size as $H$, and so containment means equality.

davidlowryduda
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