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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 ??!!!

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1 Answer 1

up vote 3 down vote accepted

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.

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thanx :) you are right :) –  Maths Lover Feb 2 '13 at 8:25

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