Apparently $g^{-1}Gg=G$ for all $g$ in $G$. I understand that by closure if you multiply on the right by $g$ and the left by $g^{-1}$ you will get an element of the group $G$. However, how do i know that some of the elements of $g^{-1}Gg$ will not be duplicate and hence I will not be able to generate the entire group $G$?
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In general, for a fixed $g \in G$, the conjugation action $x \mapsto g^{-1}xg$ is an automorphism of $G$ (i.e., a bijective homomorphism $G \rightarrow G$). So for any subgroup $H \leq G$ with $g^{-1}Hg \subseteq H$, we must actually have $g^{-1}Hg = H$. – ah11950 Apr 12 '14 at 11:01
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The bijectivity of $x\mapsto g^{-1}xg$ follows from tha fact that $x\mapsto gxg^{-1}$ is a (two-sided) inverse map. – Hagen von Eitzen Apr 12 '14 at 11:44
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Be careful also that contrary to what you seem to think, even having "duplicates" does not imply that you do not "generate the entire group", unless the group is assumed to be finite. In other words, a non-injective map from an infinite set to itself can be surjective. – fkraiem Nov 09 '14 at 12:43
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Let $h\in G$. Then for $x=ghg^{-1}$ you get $g^{-1}xg=h$. Therefor $h\in g^{-1}Gg$. Besides, remember that a subgroup $H$ is normal in $G$ if $g^{-1}Hg\subseteq H$ and that, as you noted, follows immediately for $H=G$.
Ittay Weiss
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Fix an element $x\in G$. You have to show $x\in g^{-1}Gg$. So you have to find $h\in G$ such that $x=g^{-1}hg$. Simply set $h=gxg^{-1}$.
Jack M
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let $G$ be a group. By definition
$N\lhd G $ iff $\forall a\in G, ~ aNa^{-1}\subseteq N $
but in your case $N=G$. If we show that $aGa^{-1}\subseteq G$ ,$\forall a\in G $, then we are done.
Take any arbitrary element $x$ in $aGa^{-1}$.
WTS:- $x\in G.$
let $x=aga^{-1}$ where $g,a\in G$ . So clearly , $x\in G$ . Since $G$ is a group for any $g,a\in G$, $aga^{-1}=x\in G $. Hence $aGa^{-1}\subseteq G$.
$\therefore G\lhd G.$
albo
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