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Show that if $G$ is a group and $H$ is its subgroup then $[G:H]=[G:gHg^{-1}]$, $g \in G$.

Attempted solution:

Let $f:G\mapsto\hat{G}$ be a group homomorphism such that $\mbox{Ker}f \subseteq H$ we will try to show that $[G:H]=[f(G):f(H)]$. Define a map $\phi: xH\mapsto f(x)f(H)$. Function $\phi$ is injective as $f(x)f(H)=f(y)f(H)\Rightarrow f^{-1}(y)f(x)f(H)=f(H) \Rightarrow f(y^{-1}x)=f(h_1)\Rightarrow f(y^{-1}xh_1^{-1})=1_{\hat{G}}$ for some $h_1 \in H$. Since $\mbox{Ker}f \subseteq H$ we have $y^{-1}xh_1^{-1}=h_2$ for $h_2 \in H$. Finally, $xh_1=yh_2$ and, hence, $xH=yH$.

Also, $\phi$ is well defined since $xH=yH$ implies $f(xH)=f(yH)\Rightarrow f(x)f(H)=f(y)f(H)$. Surjectivity of $\phi$ is obvious. Since $\phi$ is a bijection, $[G:H]=[f(G):f(H)]$.

To complete the proof we take $f=f_g: G\rightarrow G$, with the rule $x\mapsto gxg^{-1}$, $g\in G$.

Is it correct? Perhaps someone knows more elegant proof?

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    $\begingroup$ What you wrote after "well defined" is wrong. you have to prove that $\;xH=yH\implies f(x)H=f(y)H\;$ . What you began to prove there (more or less: it's kind of blurry at the end) is that $\;f\;$ is $\;1-1\;$ . WHat you tried to prove after "is injective" is that the map is well defined...:)...but I think you didn't succeed: that's is wrong. $\endgroup$
    – DonAntonio
    Dec 13, 2013 at 17:09
  • $\begingroup$ True. Sry. Well defined and injective confused $\endgroup$
    – VAL9000
    Dec 13, 2013 at 17:15
  • $\begingroup$ I swapped them around. But what is wrong with well definedness now? $\endgroup$
    – VAL9000
    Dec 13, 2013 at 17:19
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    $\begingroup$ check this: The function's well defined: $$xH=yH\implies y^{-1}x\in H\implies f(y)^{-1}f(x)=f(y^{-1}x)\in f(H)\implies$$ $$\phi(xH):=f(x)f(H)=f(y)f(H)=:\phi(yH)$$ $\endgroup$
    – DonAntonio
    Dec 13, 2013 at 17:26
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    $\begingroup$ But as for injectivity I can't make it. This is what I did:(2) The function's one-to-one: $$\phi(xH)=f(x)f(H)=f(y)f(H)=\phi(yH)\implies f(y^{-1}x)=$$$$=f(y)^{-1}f(x)\in f(H)\implies\ldots ??$$ $\endgroup$
    – DonAntonio
    Dec 13, 2013 at 17:26

2 Answers 2

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Hint: construct directly a set theoretic bijection (not a homomorphism) between the cosets of $H$ and those of $gHg^{-1}$.

Fix a $g \in G$ and let $\mathcal{H}=\{xH: x \in G\}$ the set of left cosets of $H$ and let $\mathcal{H}'=\{xH^g: x \in G\}$ the set of left cosets of $H^g:=gHg^{-1}$. Define $\phi:\mathcal{H} \rightarrow \mathcal{H}'$ by $\phi(xH)=x^gH^g$, where $x^g:=gxg^{-1}$. We will show that $\phi$ is well-defined, injective and surjective.

Suppose $xH=yH$, then $y^{-1}x \in H$ and hence $gy^{-1}xg^{-1} \in H^g$. But $gy^{-1}xg^{-1}=(gy^{-1}g^{-1})(gxg^{-1})= (y^{-1})^g(x^g)=(y^g)^{-1}x^g$. And this implies $x^gH^g=y^gH^g$, so $\phi$ does not depend on the particular coset representative, that is, $\phi$ is well-defined.

Now assume $\phi(xH)=\phi(yH)$, so $x^gH^g=y^gH^g$. Then $(y^g)^{-1}x^g \in H^g$. But $(y^g)^{-1}x^g=(gyg^{-1})^{-1}gxg^{-1}=gy^{-1}g^{-1}gxg{-1}=gy^{-1}xg^{-1}=(y^{-1}x)^g$. It follows that $y^{-1}x \in H$, so $xH=yH$ and $\phi$ is injective.

Finally, pick an arbitrary $xH^g \in \mathcal{H}'$. Then $\phi(x^{g^{-1}}H)=xH^g$, which means that $\phi$ is surjective.

This concludes the proof: $\#\mathcal{H}=\#\mathcal{H}'$, that is index$[G:H]=$index$[G:H^g]$.

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  • $\begingroup$ I also thought of this, yet...what function?! $\endgroup$
    – DonAntonio
    Dec 13, 2013 at 18:40
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    $\begingroup$ @DonAntonio: $xH \mapsto x^gH^g$ $\endgroup$ Dec 13, 2013 at 19:22
  • $\begingroup$ Exactly! That is the one. A well-definef bijection. $\endgroup$ Dec 13, 2013 at 19:28
  • $\begingroup$ I'm just interested in seeing the proof that that is well defined and injective, @NickyHekster ... $\endgroup$
    – DonAntonio
    Dec 13, 2013 at 19:30
  • $\begingroup$ OK will edit my answer/hint. $\endgroup$ Dec 13, 2013 at 19:43
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Elegant? Invoke orbit-stabilizer on the coset-space $G/H$, using stabilizer of $H$ and of $gH$.

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    $\begingroup$ Really nice answer, follows from the facts that action of $G$ on $G/H$ is transitive and $Stab_{G}(gH)=gHg^{-1}$. $\endgroup$
    – user371231
    Sep 24, 2021 at 9:09

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