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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!

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1 Answer 1

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.

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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

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