# $G$ a group where $H$ is a subset of $G$ with index $[G:H]<+\infty$

$G$ is a group where $H$ is a subgroup of $G$ with index $[G:H]< +\infty$. Show if for the element g is an element in G we have $gHg^{-1}$ is a subset of H. Then we must have$gHg^{-1}=H$.

(This is not necessarily true if $[G:H]= +\infty)$

Just came to this question while looking for problems. I would like to know how to prove this.

Here is what I did

for all $x^2 = e$

$$x\rightarrow x^{-1}=x$$ $$y\rightarrow y^{-1}$$ $$xy\rightarrow (xy)^{-1} = y^{-1}x^{-1}$$

Let $T_g: x\mapsto g^{-1}(x)(g)$. Let $G: g\mapsto g^{-1}xg = X$ where this is homomorphic Then $gh\rightarrow T_{gh} (x) = (gh)^{-1}xg$. Not sure how to solve. Please help.

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I removed the (measure-theory) and (general-topology) tags which were not appropriate. – Sammy Black Mar 21 '13 at 4:06
@9959: notice $H=ege^{-1}\subset gHg^-1$ and $|gHg^-1|=|H|$ – Maisam Hedyelloo Mar 21 '13 at 4:08
Where did $T$ suddenly come from. And when you write $Tsub(g)sub(h)$ do you mean $T_{g}T_h$? – Thomas Andrews Mar 21 '13 at 4:16
Do you mean $T_{gh}$ by Tsub(g)sub(h)? You can typeset it like this: $T_{gh}$. – Martin Sleziak Mar 21 '13 at 4:16
For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Mar 21 '13 at 4:17

If $[G:H]=n$ is finite, $G$ can be written as a disjoint union of $n$ cosets of $H$, $G=\bigcup_{1\le i\le n} a_i H$. Conjugating this by $g$ expresses $G$ as a disjoint union of $n$ cosets of the subgroup $gHg^{-1}$: $$G=\bigcup_{1\le i\le n} ga_i H g^{-1}=\bigcup_{1\le i\le n} (ga_ig^{-1})( g H g^{-1}),$$ so $[G:gHg^{-1}]=n$. However, if $gHg^{-1}\le H$, then also $$[G:gHg^{-1}]=[G:H][H:gHg^{-1}].$$ Cancelling $n$ from both sides then gives $[H:gHg^{-1}]=1$, so $H=gHg^{-1}$.

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Conjugating by $g$ is an automorphism of $G$. So, if $H$ is a subgroup of finite index then $gHg^{-1}$ is also a subgroup of finite index and $[G:H]=[G:gHg^{-1}]$. Therefore if $gHg^{-1}\subseteq H$, then $H=gHg^{-1}$.

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