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

Question

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

Answer 1

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

Answer 2

$\textbf{Proof:} \, \text{We know} $ $G/Z(G) \cong \text{Inn}(G)$. Since $G$ is finite, $n=|G:Z(G)|= \frac{|G|}{|Z(G)|}=|\text{Inn}(G)|$ by Lagrange's Theorem. So, there are $n$ distinct inner automorphisms $\theta_{g_i}$ where $g_i \in G$ for $i=1,...,n$. Let $x \in G$, then $\text{cl}(x)=\{g^{-1}xg | g \in G\}=\{\theta_{g_i}(x) | g_i \in G \,\, \text{for}\,\, i=1,...,n\}$. Then $|\text{cl}(x)| \leq n$.