# $[G:Z(G)] = n$ prove that each conjugacy class has at most n elements

$[G:Z(G)] = n$ prove that each conjugacy class has at most n elements.

what i tried -

I know from the orbit stabiliser theorem that

$|G| = \sum_{x_i} |G:C(x_i)| + |Z(G)|$

Because Z(G) < G i also know from Lagrange theorem that:

$|G| = |Z(G)| * [G:Z(G)] = |Z(G)| * n$

i know that the size of each conjugacy class is $|G:C(x_i)|$

here I'm stuck.. any help will be appreciated

• What is a "conjugacy group"? Conjugacy classes do not form subgroups (except the conjugacy class of identity, which is the trivial subgroup). – lisyarus Sep 7 '15 at 15:25
• $Z(G) \leq C(x_i)$ Implies that $[G : C(x_i)] \leq [G:Z(G)]$. – N. S. Sep 7 '15 at 15:25
• so Conjugacy class are not subgroups? – user2993422 Sep 7 '15 at 15:28
• N.S - why is Z(G) < C(xi)? – user2993422 Sep 7 '15 at 15:29
• $C(x_i)$ is the set of all $g\in G$ that commutes with $x_i$. On the other hand $Z(G)$ is the set of all elements that commute with everything, in particular with $x_i$. – Thibaut Dumont Sep 7 '15 at 15:47

Ok ,

so from looking at the comments -

$Z(G)≤C(x_i)$Implies that $[G:C(x_i)]≤[G:Z(G)]$

so because

$[G:Z(G)]=n$

we get -

$[G:C(x_i)]≤ n$

thanks N.S. for the help