10
$\begingroup$

Let $G$ be a finite group and $x,y \in G$ have the same order . Then is it true that there exists a group $H$ containing $G$ as a subgroup ( or an isomorphic copy) such that $x,y$ are conjugate in $H$ ? What I thought is to embed $G$ in $S_{|G|}$ , but the thing is can we show that the natural images of $x,y$ in $S_n$ ($n=|G|$ ) has same cycle structure ? If not , then is there any other way (is the claim true ?) ? Please help.

$\endgroup$
  • $\begingroup$ Your construction is the basis for Hall's universal countable locally finite group, as you can even replace $x,y$ by any finite isomorphic subgroups of $G$. $\endgroup$ – j.p. Dec 12 '16 at 8:33
6
$\begingroup$

Your approach works. Say that $x$ and $y$ have order $d$. Note that if $g\in G$, then the cycle of $g$ when $x$ acts on $G$ by translation has exactly $d$ elements, namely $g,xg,x^2g,\dots,x^{d-1}g$. So all the cycles of the image of $x$ in $S_{|G|}$ have length $d$, and so the cycle structure of this image is just $|G|/d$ $d$-cycles. The same is true of $y$, so the images of $x$ and $y$ in $S_{|G|}$ are conjugate.

(Note that the hypothesis that $G$ is finite is unnecessary. This argument doesn't quite work if $G$ is countably infinite and $x$ and $y$ have infinite order, since you don't know that $x$ and $y$ must have the same number of cycles. But you can fix this by first embedding $G$ in $G\times K$ for some infinite group $K$, to ensure $x$ and $y$ each have infinitely many cycles.)

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

What you are looking for are HNN-extensions: introduced in a 1949 paper Embedding Theorems for Groups by Graham Higman, B. H. Neumann and Hanna Neumann, it embeds a given group $G$ into another group $H$, in such a way that two given isomorphic subgroups of $G$ are conjugate (through a given isomorphism) in $H$. Applied to your situation: $\langle x \rangle$ and $\langle y \rangle$ are isomorphic.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But in the HNN-extension $G$ will not be finite and the question asks for a finite group $G$. $\endgroup$ – Derek Holt Dec 12 '16 at 17:33
  • $\begingroup$ Derek, the overgroup is $H$, and I do not see any finiteness conditions on $H$.... $\endgroup$ – Nicky Hekster Dec 12 '16 at 18:56
  • $\begingroup$ Yes you are right, I midread the question. But it is a strange question, because there is not much point in assuming that $G$ is finite if you are not going to ask for $H$ to be finite. $\endgroup$ – Derek Holt Dec 12 '16 at 21:57
  • $\begingroup$ anyway it works because $H$ is residually finite, so some finite quotient of $H$ (in which $G$ is mapped injectively) makes the job. $\endgroup$ – YCor Jun 16 '17 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy