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2

Fix the following objects:

$G$= finite group,

$x,y$ - distinct elements of $G$ of same order.

Q. Can we embded $G$ in a finite group $G_1$ such that $x,y$ become conjugate in $G_1$?

An application of HNN extension theorem perhaps does not consider embedding into finite groups for above problem (see Theorem 3.3 here). Perhaps, it simply done by adding a generator $z$ to $G$ and define a relation $z^{-1}xz=y$. But I couldn't ensure whether the embedding can be done in a finite group, provided original group is also finite.

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I don't know whether down-voters are familiar with HNN extensions, and whether they are thinking about Cayley's theorem! – p Groups 8 hours ago
    
Wht's reason for down-vote? Isn't it possible on MathStack to know down-voters reply before down-voting it? Otherwise any one can downvote any question and answer with no reason! – p Groups 8 hours ago
    
So equivalently, can we embed in a symmetric group in such a way that the two elements get the same cycle type. And we can of course assume that the elements are already in some smaller symmetric group themselves. – Tobias Kildetoft 8 hours ago
2  
I think it is part of the underlying philosophy of stackexchange that downvoters and also upvoters can remain anonymous and are not required to provide any reason. It is also often infectious - people meaninglessly copy other downvoters. So it is not worth losing any sleep over it! – Derek Holt 8 hours ago
2  
I did not downvote, but maybe someone remembered the very similar question math.stackexchange.com/q/2055099/669 from 11 days ago (can be found also among the related questions on the right, unless one sees only the not-so-great mobile version of this page). The only difference was that the requirement $G_1$ finite was given only indirectly by the chosen tags, and the accepted answer answers your question. – j.p. 7 hours ago
up vote 8 down vote accepted

Actually, after thinking a bit more on it, this turns out to follow from the proof of Cayley's theorem. By construction, the embedding in that sends an element of order $m$ to a product of disjoint $m$-cycles (since this is the action by left translation), so all elements of the same order become conjugate.

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It seems correct now, but I will think on it a little; – p Groups 8 hours ago
up vote 7 down vote

The answer is yes. As Tobias Kildetoft indicated in the comments, in the regular representation of $G$ all non-identity elements act fixed point freely. So the two elements of the same order $n$ both consist of $|G|/n$ cycles of length $n$, and hence they are conjugate in the symmetric group on the elements of $G$.

More generally, if $A$ and $B$ are isomorphic subgroups of a finite group $G$ and $\phi:A \to B$ is an isomorphism, then you can embed $G$ in a finite group $X$ with an element $g \in X$ such that $g^{-1}ag=\phi(a)$ for all $a \in A$. This is a result of B.H. Neumann - I was looking for the original reference but I haven;t found it yet - he wrote several papers on similar themes.

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Thanks for answer as well as general comments(2nd para). – p Groups 8 hours ago
    
I think it might be this article. – Andreas Caranti 5 hours ago

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