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Given two conjugate subgroups (no assumptions on finiteness), if one is yet a subgroup of the other, do they have to be identical?

Since these two subgroups are conjugate to each other, they are isomorphic; but a group can be isomorphic to its proper subgroup (e.g. additive group of integers and additive group of even integers). Could conjugacy offer a stronger argument to force equality of the two subgroups?

Thank you.

X-Naut PhD
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  • Maybe to give some intuition to why it's not enough you could think of the group as a subgroup of some $S_n$. Then 2 subgroups are conjugated iff they contain elements of the same structure written in cycle notation. But note that the structure of 2 permutations alone doesn't imply the structure of their product. Thus, at least intuitivly we can see we don't have enough info to determine isomorphism. Hope it made it clearer. – agent_cracker103 Mar 15 '23 at 22:57
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    @agent_cracker103 I doubt it. I cannot see which isomorphism you wanted to determine, and anyway in finite groups, if a subgoup is contained in one of its conjugate then they are equal. May be you did not read the question carefully? – Anne Bauval Mar 15 '23 at 23:17
  • Answers and comments kindly provided so far have made me appreciate the utility of the additional differentiable structure as I first encountered this question for Lie groups. I initially suspected that it was of purely algebraic nature, but now I realized that the dimension of the underlying differentiable manifold serves as a key invariant-- conjugation by inner automorphism is a diffeomorphism. Thank you all for your help, I really like the counter-examples, and nice to learn HNN extension. Greatly appreciated! – X-Naut PhD Mar 16 '23 at 00:48
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    I like both answers but have no idea how to accept more than one, I only upvoted both. Please know that I am very thankful for your responses! – X-Naut PhD Mar 16 '23 at 00:50

2 Answers2

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No. Consider the free group on two elements $F_2=\langle a,b\rangle$ and let

$$H=\langle a^nba^{-n}\ :\ n\geq 0\rangle$$

Now note that $aHa^{-1}$ is a proper subgroup of $H$.

This is true for finite groups though. Simply because a finite set cannot have equinumerous proper subset.

freakish
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No, because two isomorphic subgroups of a group are always conjugate in some larger group (this is the principle of HNN extensions), and you can apply this to the two subgroups $\Bbb Z$ and $2\Bbb Z$ of $\Bbb Z$ which you quoted. This gives:

In the group $$\langle s,t\mid tst^{-1}s^{-2}\rangle,$$ the two subgroups generated respectively by $s$ and by $s^2$ are conjugate but the latter is strictly contained in the former.

Anne Bauval
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  • What's the easiest way to see that $\langle s^2\rangle$ is strictly contained in $\langle s \rangle$? (To ask it a different way, how does one prove that $s$ does not have odd order?) – Jason DeVito - on hiatus Mar 15 '23 at 22:05
  • $\langle s\rangle\cong\Bbb Z$ by construction. – Anne Bauval Mar 15 '23 at 22:07
  • Sorry - I should have clarified. I had already read the Wikipedia page, and understood the Key Properties/Britton's Lemma implies what I was asking. But I was hoping there'd be an easy proof in this extra-special case. – Jason DeVito - on hiatus Mar 15 '23 at 22:08
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    Sorry, I tried but couldn't find any easier proof. – Anne Bauval Mar 15 '23 at 22:24
  • I figured out a fairly inelegant proof: first, notice that for $\mathbb{Q}$ (as an abelian group) the map $\phi(q)= 2q$ defines a group automorphism. Consider the map $\phi:\mathbb{Z}\rightarrow Aut(\mathbb{Q})$ given by $n\mapsto \phi^n$, we can form the semidirect product $G:=\mathbb{Q}\rtimes \mathbb{Z}$. Now, if I haven't miscalculated, the map $a\mapsto (1,0)$ $b\mapsto (0,1)$ defines a map from the free group on $\langle a,b\rangle$ to $G$ which has $bab^{-1}a^{-2}$ in its kernel. This descends to your group, where $s$ maps to $(1,0)]\in G$, which has infinite order. – Jason DeVito - on hiatus Mar 16 '23 at 02:08
  • Applause! ;-) . – Anne Bauval Mar 16 '23 at 06:17