Abstract Algebra – Order of a Conjugacy Class Bounded by 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)| $

$|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 positive integar 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 $| G : C_G (g_i)|\le n$ 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.