prove that $[G: xHx^{-1}]=[G:H]$ Let $G$ be a group, $H$ is a subgroup of $G$ and $x\in G$, prove that $[G: xHx^{-1}]=[G:H]$.
I have proved that $xHx^{-1}\leq G$ and $|xHx^{-1}|=|H|$. I tried to construct a map from $\left\{a xHx^{-1}:a\in G\right\}$ to $\left\{aH:a\in G\right\}$, but I have difficulty in proving it's injective. I am new on abstract algebra, please do not use Lagrange Theorem.
Thanks!
 A: If you are dealing with finite groups you are already done. The definition of $[G:H]$ is "how many times the subgroup $H$ stay in $G$?" That is $|G|/|H|$ (which is an integer by Lagrange). Since you have proved that $|x^{-1}Hx|=|H|$ then it's done.
A: Allow me to indicate what the general line of reasoning that should be used to prove this result.

*

*Quotient maps induced by maps compatible with equivalence relations. Consider a map $f \colon A \to B$ and the canonical equivalence it induces on $A$, denoted by $\mathrm{Eq}(f)\colon=\{(x, y)|\ x, y \in A \wedge f(x)=f(y)\}$. Given an equivalence $R$ on $A$ such that $R \subseteq \mathrm{Eq}(f)$, there exists a unique map $g \colon A/R \to B$ such that $f=g \circ s$, where $s \colon A \to A/R$ denotes the canonical surjection. $g$ is called the quotient of $f$ with respect to $R$. It is easy to prove that $g$ is surjective iff $f$ is surjective and that $\mathrm{Eq}(g)=(s \times s)[\mathrm{Eq}(f)]$ (which entails the equivalence $g$ injective iff $\mathrm{Eq}(f)=R$).

*Particular version of quotient maps. Consider now a map $f \colon A \to B$ where $A$ and $B$ are each equipped with an equivalence relation $R$ and $S$ such that $(f \times f)[R] \subseteq S$. Then there exists a unique map $h \colon A/R \to B/S$ such that $q \circ f=h \circ p$, where $p \colon A \to A/R$ respectively $q \colon B \to B/S$ denote the canonical surjections. It can be easily shown that $\mathrm{Eq}(h)=(p \times p)\left[(f \times f)^{-1}[S]\right]$ -- to the effect that $h$ is injective iff $R=(f \times f)^{-1}[S]$ -- and that $h$ is surjective iff $f[A]$ forms a complete (not necessarily independent) system of representatives of $S$ on $B$. The map $h$ is also called the quotient of $f$ with respect to $R$ and $S$.

*Further particularisation of 2). If in the above setting $f$ is bijective such that $(f \times f)[R]=S$ it follows easily that the quotient map $h$ is also bijective (its inverse will be none other than the quotient of the inverse of $f$ with respect to $S$ and $R$, notice the reversed order).

*Let us now consider the particular case of a group isomorphism $f \colon G \to G'$. If $H \leqslant G$ is an arbitrary subgroup, let $H'\colon=f[H]$ and $R$, $S$ denote the left congruences on $G$ modulo $H$ respectively on $G'$ modulo $H'$. It follows right away from the definition of a left congruence and the hypothesis of bijectivity on $f$ that $(f \times f)[R]=S$, so that by virtue of 3) there exists a bijection $h \colon (G/H)_{\mathrm{s}} \to \left(G'/H'\right)_{\mathrm{s}}$, namely the quotient of $f$ with respect to $R$ and $S$. By definition of the index of a subgroup, we infer that $(G \colon H)=\left|(G/H)_{\mathrm{s}}\right|=\left|\left(G'/H'\right)_{\mathrm{s}}\right|=(G' \colon H')$.

This applies in particular to your case, where the isomorphism $f$ is the inner automorphism of $G$ given by left conjugation with $x$.
