Finding the number of conjugacy classes

**kimberu** · April 4th 2010, 08:36 PM

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

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

thanks!

**NonCommAlg** · April 4th 2010, 11:21 PM

Originally Posted by kimberu

Given a group G, non-abelian, with |G|= $p^3$ , show G has $p^2 + p -1$ 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 $Z(G)$ be the center of $G$ and $C(g)$ the centralizer of $g \in G$ in $G.$ see that $|Z(G)|=p$ and also $|C(g)|=p^2,$ for all $g \notin Z(G),$ because $Z(G) \subset C(g).$ now apply the class equation. $\Box$

**kimberu** · April 5th 2010, 12:06 AM

So the class equation says:

G

=

Z(G)

+

[G:C( $x_i$ )]

meaning,
$p^3=p +$ 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!)

**NonCommAlg** · April 5th 2010, 12:13 AM

Originally Posted by kimberu

So the class equation says:

G

=

Z(G)

+

[G:C( $x_i$ )]

meaning,
$p^3=p +$ 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 $x_i$ in your sum in the right hand side are not in $Z(G)$ and thus, as i mentioned, $[G:C(x_i)]=\frac{|G|}{|C(x_i)|}=p.$ now suppose $k$ is the number of conjugacy classes. then:

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

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