The order of a conjugacy class is bounded by the index of the center

Question

If the center of a group $G$ is of index $n$, prove that every conjugacy class has at most $n$ elements. (This question is from Dummit and Foote, page 130, 3rd edition.)

Here is my attempt: we have $$ |G| = |C_G (g_i)| |G : C_G (g_i)| \,, \quad |C_G (g_i)| = |Z(G)| \cdot |C_G (g_i):Z(G)| \,. $$ Then $$ |G| = |Z(G)| \cdot |C_G (g_i):Z(G)| \cdot |G : C_G (g_i)| \,. $$ Then $$ n = |C_G (g_i):Z(G)| \cdot |G : C_G (g_i)| \,. $$

But $|C_G (g_i):Z(G)|$ is a positive integer as $Z(G)$ is subgroup of $C_G (g_i)$.

If $|G : C_G (g_i)|$ bigger than $n$, then $|C_G (g_i):Z(G)| \cdot |G : C_G (g_i)|$ is bigger than $n$.

But $|C_G (g_i):Z(G)| \cdot |G : C_G (g_i)| = n$, contradiction.

Then $|G : C_G (g_i)| \leq n$ and this completes the proof.

Is this proof right or not?

Answer 1

Your proof is correct, but you can simplify the end by saying that $$[G:C_G(g_i)] \leq [C_G(g_i):Z(G)]\cdot[G:C_G(g_i)] =n$$ Actually a contradiction is not needed.

Answer 2

From the work you did you can also prove the slightly stonger fact, that: $$ |x^G|\Big|[G:Z(G)] $$ We know that $Z(G)\leq C_G(x)\leq G$ and $|x^G|=[G:C_G(x)]$ so since the index is multplicative (if $H\leq K\leq G$ then $[G:H]=[K:H][G:K]$) we get that $$[G:Z(G)]=[C_G(x):Z(G)][G:C_G(x)]$$ And we get the result