Subgroup and conjugacy class let $G$ be a group and $C_x$ a conjugacy class, with $|C_x|=n$. prove that $\exists H\leq G$ with H being a subgroup of G, that $|G/H|=n$
It easy to proof that this happens with G being a finite group but my problem starts when G is a infinite group because I cannot apply Lagrange theorem and because of that I don't have any ideia of who is H
Any hints?
 A: Consider
$F_x = \{f \in G, \; fxf^{-1} = x \}; \tag 1$
that is, $F_x \subset G$ is the set of group elements which fix $x \in G$ under conjugation; it is easy to see that $F_x$ is in fact a subgroup of $G$, since for 
$a, b \in F_x \tag 2$
we have
$(ab)x(ab)^{-1} = (ab)x(b^{-1}a^{-1}) = a(bxb^{-1})a^{-1} = axa^{-1} = a, \tag 3$
and the identity $e \in G$ is clearly in $F_x$:
$exe^{-1} = exe = xe = x; \tag 4$
and
$a \in F_x \Longleftrightarrow axa^{-1} = x \Longleftrightarrow a^{-1}xa = x \Longleftrightarrow a^{-1} \in F_x. \tag 5$
Now consider any coset $gF_x$ of $F_x$, where $g \in G$; for $f \in F_x$ we have
$(gf)x(gf)^{-1} = (gf)x(f^{-1}g^{-1}) = g(fxg^{-1})g^{-1} = gxg^{-1}, \tag 6$
which shows that elements $gf \in gF_x$ all take $x$ to $gxg^{-1}$ under conjugation; in fact, the conjugate $gxg^{-1}$ only depends on the coset $gF_x$ and not upon its representative $g$; for if
$g_1F_x = g_2F_x, \tag 7$
then
$g_1 = g_1e = g_2 f_1 \tag 8$
for some $f_1 \in F_x$, whence
$g_1xg_1^{-1} = (g_2f_1)x(g_2f_1)^{-1} = (g_2f_1)x(f_1^{-1}g_2^{-1}) = g_2(f_1xf_1^{-1})g_2^{-1} = g_2xg_2^{-1}; \tag 9$
it follows then, that there is a well-defined map
$\phi: G/F_x \to C_x; \tag{10}$
$\phi$ is injective, for if
$\phi(g_1F_x) = \phi(g_2F_x), \tag{11}$
then
$g_1xg_1^{-1} = g_2xg_2^{-1} \Longrightarrow (g_2^{-1}g_1)x(g_1^{-1}g_2) = x \Longrightarrow (g_2^{-1}g_1)x(g_2^{-1}g_1)^{-1} = x$
$\Longrightarrow g_2^{-1}g_1 \in F_x \Longrightarrow g_1 = g_2f, \; f \in F_x \Longrightarrow g_1F_x = g_2fF_x = g_2F_x; \tag{12}$
$\phi$ is also surjective, for
$\phi(gF_x) = gxg^{-1} \tag{13}$
for any conjugate of $x$.  Since $\phi$ is a bijection, we conclude that
$\vert G/F_x \vert = \vert C_x \vert = n, \tag{14}$
$OE\Delta$.
A: Hint:  Notice that $C_x$ is an orbit of $x$ under the conjugation action of $G$ (on its own elements).  Some elements of $G$ stabilize $x$ when they conjugate $x$ and some elements of $G$ push $x$ to a different member of this orbit.
