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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}$.

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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): math.stackexchange.com/questions/72036/… –  Mikko Korhonen Jan 14 '12 at 22:27
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@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

1 Answer 1

up vote 10 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 $p$.

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