# Thread: Finding the number of conjugacy classes

1. ## Finding the number of conjugacy classes

Given a group G, non-abelian, with |G|=$\displaystyle p^3$, show G has $\displaystyle p^2 + p -1$ distinct conjugacy classes.

Any help would be great -- I'm sure there's a useful theorem or fact I'm missing out on.

thanks!

2. Originally Posted by kimberu
Given a group G, non-abelian, with |G|=$\displaystyle p^3$, show G has $\displaystyle p^2 + p -1$ distinct conjugacy classes.

Any help would be great -- I'm sure there's a useful theorem or fact I'm missing out on.

thanks!
let $\displaystyle Z(G)$ be the center of $\displaystyle G$ and $\displaystyle C(g)$ the centralizer of $\displaystyle g \in G$ in $\displaystyle G.$ see that $\displaystyle |Z(G)|=p$ and also $\displaystyle |C(g)|=p^2,$ for all $\displaystyle g \notin Z(G),$ because $\displaystyle Z(G) \subset C(g).$ now apply the class equation. $\displaystyle \Box$

3. So the class equation says:
G=Z(G)+[G:C($\displaystyle x_i$)]

meaning,
$\displaystyle p^3=p +$ this sum, but what exactly is the sum? Is it just p, m times? (sorry, I've never used this equation before. thanks so much for the help!)

4. Originally Posted by kimberu
So the class equation says:
G=Z(G)+[G:C($\displaystyle x_i$)]

meaning,
$\displaystyle p^3=p +$ this sum, but what exactly is the sum? Is it just p, m times? (sorry, I've never used this equation before. thanks so much for the help!)
the $\displaystyle x_i$ in your sum in the right hand side are not in $\displaystyle Z(G)$ and thus, as i mentioned, $\displaystyle [G:C(x_i)]=\frac{|G|}{|C(x_i)|}=p.$ now suppose $\displaystyle k$ is the number of conjugacy classes. then:

$\displaystyle p^3=p + (k - |Z(G)|)p=p+(k-p)p,$ which will give us the result.