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?
Best Answer
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$[G:H^g]$.