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let $G$ be a group. Is it true that the only subgroup of $G$ that is conjugate to $G$ is $G$ itself? if $G$ is finite this is clear as conjugate subgroups have the same order but what about infinite groups?

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Assume that you conjugate with an element $x\in G$. Can you write an arbitrary element $y\in G$ in the form $y=x^{-1}zx$ for some $z\in G$? –  Jyrki Lahtonen Jul 2 '11 at 8:30
    
@Jyrki Lahtonen: of course $y=x^{-1}(xy{x^-1})x$ but my question is why $G$ can't be conjuguate to another subgroup –  palio Jul 2 '11 at 8:41
    
If G is conjugate to H, so that H is the image of G under conjugation b $x$, and H is a proper subgroup of G, take $y$ in G\H. You've just proved that $y$ is in the image, so H cannot be proper. –  Mark Bennet Jul 2 '11 at 9:00
    
@Mark Bennet: it is $xyx^{-1}$ that is in the image $H$ not $y$ –  palio Jul 2 '11 at 9:14

1 Answer 1

up vote 2 down vote accepted

Each conjugation induces a permutation of $G$ as a set. When you permute a set, proper subsets remain proper subsets.

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you mean every $g_0\in G$ induces a permutation $G\rightarrow G;\, g\mapsto g_0gg_0^{-1}$ then what? –  palio Jul 2 '11 at 8:54
    
That's what I mean. Then if $H$ is a proper subgroup of $G$, in particular it's a proper subset, so the permutation sends it to another proper subset. That proper subset can't be $G$ itself. –  user92843 Jul 2 '11 at 8:57
    
you mean the restriction of the permutation to a subgroup $H$ of $G$ is never surjective? it's clear why it is never surjective by order argument but for infinite $G$ it is not clear for me –  palio Jul 2 '11 at 9:04
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This is really a fact about sets and not groups, as I explicitly stated in my answer. If $\pi$ is a permutation of a set $X$, and $Y$ is a proper subset of $X$ (that is, there is some element of $X$ that is not in $Y$), then the pointwise image $\pi[Y]$ is also a proper subset of $X$. –  user92843 Jul 2 '11 at 9:07
    
thanks alot!! it's clear now –  palio Jul 2 '11 at 9:09

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