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

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
