7
$\begingroup$

Let $G$ be a non-abelian group of order $p^3$. How many are its conjugacy classes?

The conjugacy classes are the orbits of $G$ under conjugation of $G$ by itself. Since $G$ is non-abelian, its center has order $p$. So the class equation yields $p^3 = p + \sum_{[x]} (G: G_x)$, where $G_x$ is the centralizer of $x$ and the sum is taken over disjoint orbits $[x]$. We can also see that $(G:G_x)$ can only be $p$ or $p^2$. So we will have $p$ orbits of length $1$ and then orbits of length $p$ and $p^2$. Any hints on determining the number of the latter?

$\endgroup$
9
$\begingroup$

We know that $Z(G) = p$. Suppose $[G:G_x] = p^2$ for some $x \notin Z(G)$. Then $G_x$ has $p$ elements, and since $Z(G) \subseteq G_x$, we have $Z(G) = G_x$. Now since $x$ is in $G_x$, it is also in $Z(G)$. From this it follows that $G_x = G$, and thus $Z(G) = G$, a contradiction. Thus $[G:G_x] = p$ for all $x \notin Z(G)$, showing that $G$ has $p^2 + p - 1$ conjugacy classes.

$\endgroup$
  • $\begingroup$ Why not just say that the centralizer of any non-cntral element must properly contain the center? It is a little faster in the second part. $\endgroup$ – awllower Jun 3 '12 at 17:57
7
$\begingroup$

Suppose $g$ is a noncentral element. Then the centralizer of $g$ contains the centre of $G$, and also contains $g$ itself, giving it at least $p+1$ elements. Thus the centralizer has order $p^2$, so $g$ has $p$ conjugates. There are thus $p+\frac{p^3-p}{p}=p^2+p-1$ conjugacy classes.

$\endgroup$
5
$\begingroup$

The centraliser of a non-central element contains at least $p+1$ elements, since it contains the centre, with order $p$. It cannot have order equal to $p^3$, or the element would be central. Thus, the order of the centraliser or a non-central element is equal to $p^2$, and hence the index is equal to $p$, which is to say that it has $p$ conjugates. Therefore, there are $\frac{p^3 - p}{p} = p^2 - 1$ conjugacy classes of non-central elements, giving a total of $p^2 + p - 1$ conjugacy classes.

$\endgroup$
3
$\begingroup$

One can break it up into 3 steps. First, show that $|Z(G)|$ is equal to p. Second, show that if $a$ is not in $Z(G)$ then $|N_G(a)|=p^2$. Finally, the last step is to use the class equation to show that there are $p^2+p-1$ conjugacy classes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.