Index of conjugate subgroup equals the index of subgroup 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?

Elegant? Invoke orbit-stabilizer on the coset-space $G/H$, using stabilizer of $H$ and of $gH$.
