Otdfbt

let $G$ be a group and $C_x$ a conjugacy class, with $|C_x|=n$. prove that $exists Hleq G$ 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?

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

$(ab)x(ab)^-1 = (ab)x(b^-1a^-1) = a(bxb^-1)a^-1 = axa^-1 = a, tag 3$

and the identity $e in G$ is clearly in $F_x$:

$exe^-1 = exe = xe = x; tag 4$

and

$a in F_x Longleftrightarrow axa^-1 = x Longleftrightarrow a^-1xa = x Longleftrightarrow a^-1 in F_x. tag 5$

Now consider any coset $gF_x$ of $F_x$, where $g in G$; for $f in F_x$ we have

$(gf)x(gf)^-1 = (gf)x(f^-1g^-1) = g(fxg^-1)g^-1 = gxg^-1, tag 6$

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

$g_1xg_1^-1 = (g_2f_1)x(g_2f_1)^-1 = (g_2f_1)x(f_1^-1g_2^-1) = g_2(f_1xf_1^-1)g_2^-1 = g_2xg_2^-1; tag 9$

it follows then, that there is a well-defined map

$phi: G/F_x to C_x; tag10$

$phi$ is injective, for if

$phi(g_1F_x) = phi(g_2F_x), tag11$

then

$g_1xg_1^-1 = g_2xg_2^-1 Longrightarrow (g_2^-1g_1)x(g_1^-1g_2) = x Longrightarrow (g_2^-1g_1)x(g_2^-1g_1)^-1 = x$
$Longrightarrow g_2^-1g_1 in F_x Longrightarrow g_1 = g_2f, ; f in F_x Longrightarrow g_1F_x = g_2fF_x = g_2F_x; tag12$

$phi$ is also surjective, for

$phi(gF_x) = gxg^-1 tag13$

for any conjugate of $x$. Since $phi$ is a bijection, we conclude that

$vert G/F_x vert = vert C_x vert = n, tag14$

$OEDelta$.

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.