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Let $G$ be a group and $\varnothing ≠S \subseteq G$ and given an element $x \in G$ I want to prove if the following implication holds (or disprove, if it doesn't):

$xSx^{-1} \subseteq S \iff xSx^{-1} = S$

My attempt: ($\Leftarrow$) is trivial, so I'll try to prove ($\Rightarrow$).

It is sufficient to show that $S \subseteq xSx^{-1}$, so I'll try prove by contradiction. Assume $S \nsubseteq xSx^{-1} \Rightarrow \exists s \in S: s \notin xSx^{-1}$ and hence $xsx^{-1} \in xSx^{-1} \subseteq S$, so $xsx^{-1} =s'$ for some $s' ∈ S \Rightarrow s =x^{-1}s'x \Rightarrow s \in x^{-1}Sx$

But now I'm not sure how to produce a contradiction. Any help?

Aitor Iribar Lopez
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William
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    It’s obviously true for finite $S,$ but it isn’t true in general for infinite $S.$ Consider the group $\Sigma_{\mathbb N},$ the group of bijections on $\mathbb N.$ Then the subgroup that fixes all even numbers is a conjugate of the subgroup that fixes all even numbers $\neq0.$ – Thomas Andrews Jan 11 '22 at 22:37
  • @Shaun No actually, I have $s \notin xSx^{-1}$ but later I have shown that $s \in x^{-1}Sx$. Does this mean I have to establish $xSx^{-1}= x^{-1}Sx$? Also, why less use of $\implies$? Is my usage of implication incorrect? – William Jan 11 '22 at 22:39
  • @ThomasAndrews Yes, it's incomplete. I say that "I don't know how to produce a contradiction" from there in the last line. I don't think the working is incorrect though? Is it? – William Jan 11 '22 at 22:42
  • @ThomasAndrews My motivation for this question is normal subgroup, where for any subgroup $H$, $xHx^{-1} \subseteq H \iff xHx^{-1}=H$. So I was wondering if it would work for any subset $S$ as well. I'm not sure I understand "why" exactly does it work for one and not for the other. It's definitely not obvious to me. :( – William Jan 11 '22 at 22:45
  • I noticed that mistake, too, @William. That's why I deleted my answer. Thank you anyway. – Shaun Jan 11 '22 at 22:48
  • The normal subgroup has $xHx^{-1}\subseteq H$ for every $x.$ Being true for all $x$ means all $xHx^{-1}=H.$ is Being true for just some $x$ is not enough to deduce equality. – Thomas Andrews Jan 11 '22 at 23:15
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    However, if for some $x,$ $xSx^{-1}\subseteq S$ and $x^{-1}Sx\subseteq S,$ then all three are equal, $xSx^{-1}=x^{-1}Sx=S.$ – Thomas Andrews Jan 11 '22 at 23:23
  • A counterexample can be found here – Arturo Magidin Jan 12 '22 at 03:55

2 Answers2

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Counterexamples can be constructed with the concept of an "ascending HNN extension". The general HNN extension is defined by taking a group $G$, two subgroups $A,B < G$, and an isomorphism $f : A \to B$, choosing a presentation for $G$, and defining $G *_f$ to be the group defined by starting with the presentation for $G$, adding one generator $t$ called the "stable letter", and adding an extra defining relation $t a t^{-1} = f(a)$ for each $a \in A$. One proves that the natural homomorphism $G \mapsto G*_f$ is injective on $G$ and so also on $A$ and $B$.

In the group $G*_f$ it follows that $A$ is conjugate to $B$, because $tAt^{-1}=B$.

So to get a counterexample that disproves your implication, we simply need a group $G$ and two subgroups $A,B$ such that $A$ is isomorphic to one of its own proper subgroups $B < A$. In this situation the HNN extension that one gets by choosing an isomorphism $f : A \to B$ is called an ascending HNN extension.

Perhaps the very easiest example is $G = A = \mathbb Z$ and $B = 2 \mathbb Z$, with isomorphism $f : \mathbb Z \to 2 \mathbb Z$ defined by $f(1)=2$. One obtains the famous Baumslag-Solitar group $$\langle a,t \mid tat^{-1}=a^2\rangle $$ in which the infinite cyclic subgroup $\langle a \rangle$ is conjugate to its own index 2 subgroup $\langle a^2 \rangle$.

Lee Mosher
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  1. It is possible to have an automorphism $\phi$ of $G$ and a subgroup $S$ such that $\phi(S) \subset S$, but $\phi(S) \ne S$.

  2. Given a group $G$, there exists a larger group $\tilde G$ ( the holomorph of $G$) that contains $G$ and $\phi$, and such that the action of $\phi$ on $G$ is given by conjugation ( see link for details). In fact, we can use even a subgroup of the holomorph.

  3. An example in 1. is $G=\mathbb{Q}$, $S= \mathbb{Z}$, and $\phi(x) = 2x$. Now the holomorph of $\mathbb{Q}$ is the group of the affine transformation of the rational line $$f_{b,a}\colon x \mapsto a x + b$$ that can also be written as $$\mathbb{Q} \rtimes \mathbb{Q}^{\times}$$ Inside this group we have $S = \mathbb{Z} \rtimes{1}$, the element $x=(0, 2)$, and $x S x^{-1} = 2 \mathbb{Z}$.

orangeskid
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