TITLE: The order of a conjugacy class is bounded by the index of the center
QUESTION [7 upvotes]: 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 ??!!!

REPLY [5 votes]: 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.