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Let $H \le G\le F$ be a chain of groups and $f\in F$ be a fixed element with $H' := f H f^{-1} \le G$.

My question: Do we have $$ [G : H] < \infty \quad\Leftrightarrow\quad [G : H'] < \infty $$ or even $$ [G : H] = [G : H']?$$ Here I assume $[G : H]$ to be a cardinal number. I tried to prove the latter by constructing a bijection $\varphi : G/H \to G/H'$. There are two "natural" approaches to do so:

  • $gH \mapsto g f H f^{-1} = gH'$
  • $gH \mapsto f gH f^{-1}$

The problem is that the first one is not well defined while the second one doesn't map to $G/H'$. So is this enough evidence to don't bother with proving anymore but start looking for counter examples?

principal-ideal-domain
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    The last one: $fgHf^{-1}=fgf^{-1}fHf^{-1}$, so it is correct. – Nicky Hekster May 12 '17 at 08:41
  • @NickyHekster No, since we don't know if $fgf^{-1} \in G$. – principal-ideal-domain May 12 '17 at 09:09
  • Yes, you are right I was biased by the letter $G$, always indicating the overgroup. Apologies, it is then not true, that is, my suggestion and your question. You might want to read https://math.stackexchange.com/questions/107862/conjugate-subgroup-strictly-contained-in-the-initial-subgroup – Nicky Hekster May 12 '17 at 11:48

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For a counterexample, let $G$ be any group with a subgroup $H$ of infinite index with $G \cong H$. For example we could take $G = \langle a,b \rangle$ free and $H = \langle a^2,b^2 \rangle$. Let $\phi:G \to H$ be an isomorphism.

Then we can define an HNN extension $F = \langle G,f \rangle$, where $t^{-1}gt = \phi(t)$ for all $g \in G$. By the general theory of HNN extensions, $F$ contains $G$ as a subgroup, and $G$ and $H$ are conjugate in $F$.

Derek Holt
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