Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove: If $G$ is a nonabelian group with $[G : Z(G)] = n$, then every conjugacy class of $G$ has strictly fewer than $n$ elements.

My approach so far:

Observe that $Z(G) \leq C_G(a) = G_a \leq G$. So we know $$[G:Z(G)] = [G:G_a][G_a:Z(G)] = \vert C_a \vert [G_a : Z(G)]$$ and thus we can conclude that $\vert C_a \vert = [G:G_a] \mid [G:Z(G)] = n$.

Now, suppose to the contrary that $\vert C_a \vert = n$. [Try to contradict $G$ being nonabelian.]

I'm having trouble finding a contradiction to the case of $n$. Any hints, approaches? Thanks!

share|cite|improve this question
up vote 2 down vote accepted

Let $a \in G$ with $|C_a|=n$, then you must have $Z(G)=G_a$. Hence $a \in Z(G)$ and this implies $|C_a|=n=1$. We conclude that $G=Z(G)$, so $G$ must be abelian, a contradiction.

share|cite|improve this answer
So since $n = 1$, it follows that $[G:Z(G)] = 1$ giving $G = Z(G)$ giving us the abelian property. Thanks! – Robert Cardona Oct 29 '12 at 11:25
Yes, exactly. And as you already concluded the cardinality of a conjugacy class must (strictly) divide $n$. – Nicky Hekster Oct 29 '12 at 11:27
I wrote all of that down in my notebook but never made the connection of the order of the orbit causes the index of $Z(G)$ to be 1. Thanks a lot! – Robert Cardona Oct 29 '12 at 11:28
+1 Very simple way. I like it. – Babak S. Oct 29 '12 at 11:46
Can't believe I missed the $a\in Z(G)$ step. – peoplepower Oct 29 '12 at 22:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.