Exercise 4.3.5(If $|G:Z(G)| = n$, then every conjugacy class has at most $n$ elements)

Question

If the center of $G$ is of index $n$, prove that every conjugacy class has at most $n$ elements.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $a$ be any element of $G$ and let ${\mathcal O}_a$ be the conjugacy class of $a$. Since $Z \leq C_G(a)$, then
$$n = |G : Z| = |G : C_G(a)| \cdot |C_G(a) : Z|.$$
Then
$$|{\mathcal O}_a| = |G : C_G(a)| = \frac{|G : Z|}{|C_G(a) : Z|} \leq |G:Z| = n,$$
so every conjugacy class has at most $n$ elements.

\end{proof}

Exercise 4.3.5(If $|G:Z(G)| = n$, then every conjugacy class has at most $n$ elements)

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Exercise 4.3.5(If $|G:Z(G)| = n$, then every conjugacy class has at most $n$ elements)

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