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1) A theorem of Higman-Neumann-Neumann implies that any group $G$ can be embedded in a group $\tilde{G}$ such that any two isomorphic subgroups of $G$ are conjugate in $\tilde{G}$.

2) It is well known that the infinite cyclic group has only two automorphisms: identity and inversion.

Let us apply (1) for the infinite cyclic group $\langle x\rangle \cong \mathbb{Z}$. Then $\langle x\rangle$ and $\langle x^2\rangle$ are conjugate in some group $\tilde{G}$. Hence there is some $g\in \tilde{G}$ such that $g^{-1}xg=x^{2i}$ for some $i\neq 0$. This means, $g$ normalizes the subgroup $\langle x\rangle$, and induces an automorphism of $\langle x\rangle$, which takes $x$ to $x^{2i}$, which is different from $x,x^{-1}$.

I arrived at some wrong conclusion by (2).

Please clarify the argument, and theorems, if anything wrong is there.

Thanks in advance!

anon
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Beginner
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  • $g^{-1}Hg\color{Red}{\subset}H$ is what you've shown with $H=\langle x\rangle$, but that is not sufficient for $g$ to normalize $H$ (i.e. this does not logically imply $g^{-1}Hg\color{Red}{=} H$). My answer here also covers this example. – anon Aug 09 '13 at 06:30
  • $H$ is normal in $G$ if for all $h\in H$, for all $g\in G$, $g^{-1}hg\in H$ (see http://en.wikipedia.org/wiki/Normal_subgroup). This will imply, in question, that $g$ normalizes $\langle x\rangle$, isn't it? – Beginner Aug 09 '13 at 06:34
  • Why would it? ${}$ – anon Aug 09 '13 at 06:38
  • just be definition of normal subgroup! – Beginner Aug 09 '13 at 06:39
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    Hint: Think about what happens if conjugate by $g^{-1}$ instead. Will this be in the subgroup? – Tobias Kildetoft Aug 09 '13 at 06:39
  • @Tobias: got it! – Beginner Aug 09 '13 at 06:40
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    There are two definitions of two different things in play here: $H$ being a normal subgroup of $G$, and an element $g\in G$ normalizing $H$. You're quoting the first definition, but you need to be paying attention to the second definition. – anon Aug 09 '13 at 06:40
  • The concept of one element normalizing a subgroup is not as useful when the groups are not finite. You need to consider subgroups normalizing other subgroups. And the subgroup generated by $g$ will not normalize the given one. – Tobias Kildetoft Aug 09 '13 at 06:40
  • @Tobias: why not useful? – Beginner Aug 09 '13 at 06:41
  • Because of a situation as the one illustrated here. Just because the element $g$ normalizes the subgroup $H$, this does not mean it induces an automorphism of $H$ (just an injective endomorphism). – Tobias Kildetoft Aug 09 '13 at 06:50
  • @TobiasKildetoft Are you suggesting that $gHg^{-1}\subsetneq H$ would imply $g$ "normalizes" $H$? That contradicts the definition of "normalizer" on Wikipedia, Mathworld, planethmath, groupprops, and all references cited therein... – anon Aug 09 '13 at 07:23
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    @anon I guess this depends on what definition one uses of normalizes when it comes to elements. – Tobias Kildetoft Aug 09 '13 at 07:25
  • In the problem posted, $g^{-1}\langle x\rangle g\subseteq \langle x\rangle$, but $g$ is not inducing automorphism of $\langle x\rangle$. – Beginner Aug 09 '13 at 11:11

1 Answers1

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Indeed, it is possible for a subgroup to properly contain one of its conjugates. That is, it is perfectly possible for $gHg^{-1}\subsetneq H$. The normalizer $N_G(H)$ is the set of all $g$ such that $gH=Hg$, which is a stirctly stronger property than $gHg^{-1}\subseteq H$ in general (with say finite groups they are equivalent though, because conjugation is injective and we can check orders etc). Thus $gHg^{-1}\subset H$ does not tell us that $g$ normalizes $H$, and conjugation by $g$ does not restrict to an automorphism of $H$.

For the record, I take "$g$ normalizes $H$" to be equivalent to $g\in N_G(H)$, where $N_G(H)$ is the normalizer of $H$ in $G$ defined as the set of all $g\in G$ such that $gH=Hg$ (strict equality); this definition is consonant with that listed in GP, WP, MW, PM and presumably all references cited therein. I haven't seen $gHg^{-1}\subseteq H$ as the definition for "$g$ normalizes $H$" outside of any context that wasn't explicitly specifically for finite groups, if I recall correctly.

anon
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