Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

This is a problem from Preliminary Exam - Spring 1984, UC Berkeley

For a $p$-group of order $p^4 $, assume the center of $G$ has order $p^2 $. Determine the number of conjugacy classes of $G$.

What I have tried: each element of the center constitutes a conjugacy class; the other conjugacy classes have order a power of $p$; their sum is $ \ p^{4} - p^{2}$.

share|cite|improve this question
If $g \in G$, the conjugacy class of $g$ is of order a power of $p$ because it's the order of $G$ divided by the order of the centralizer of $g$. And since the centralizer of $g$ contains the center, we know the conjugacy class of $g$ is either of order $1$, $p$ or $p^2$. This should reduce the work to be done (you've already worked out the case of order $1$). – Joel Cohen Jan 14 '12 at 22:16
Thanks I have edited – WLOG Jan 14 '12 at 22:18
Notice that, in fact, no conjugacy class of $G$ has size $p^2.$ Can you see why this is? – Geoff Robinson Jan 14 '12 at 22:23
By the way, I think this book includes solutions to the problems. This question is related (and the solution is similar):… – Mikko Korhonen Jan 14 '12 at 22:27
@m.k.: you didn't say which book, but you probably meant the de Souza and Nuno-Silva book. In fact, it doesn't always contain all the solutions. – Arturo Magidin Jan 14 '12 at 22:35
up vote 11 down vote accepted

Let $K$ be a conjugacy class with more than one element. Since the order of $K$ divides $p^4$, it must be $p$, $p^2$ or $p^3$. Now $|K| = [G : C_G(g)]$, where $C_G(g)$ is the centralizer of some $g \in K$.

If $|K| = p^3$, then $|C_G(g)| = p$. This is not possible, since the center is always contained in the centralizer.

If $|K| = p^2$, then $|C_G(g)| = p^2$. Since the center is contained in $C_G(g)$, we get $C_G(g) = Z(G)$. Thus $g \in Z(G)$, implying $C_G(g) = G$ which is a contradiction.

Thus any conjugacy class with more than one element has exactly $p$ elements. Now use the class equation to find out the number of conjugacy classes in terms of $p$.

share|cite|improve this answer

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.