I came with this proof and I found some other proofs online, but mine is different and I want to see if I made any mistakes.


Suppose $|G| = p^3$, where $p$ is a prime. Show that either $|Z(G)|=p$ or $G$ is abelian.

Case 1: $G$ is not abelian

We have the class equation

$$|G| = |Z(G)| + \sum_{g\in G\setminus Z(G) } \frac{|G|}{|\mathrm{Cent}(g)|}$$

where $Z(G)$ is the center of the group, and $|\mathrm{Cent} (g)|$ is the centralizer of $g$.

if we solve for $|Z(G)|$ we get

$$|Z(G)| = |G| - \sum_{g\in G\setminus Z(G) } \frac{|G|}{|\mathrm{Cent}(g)|} = p^3 \bigg( 1-\sum \frac{1}{\mathrm{Cent}(g)} \bigg)$$

Also, since $Z(G) \leq G$, by Lagrange's theorem, $|Z(G)|$ divides $|G| = p^3 = ppp$

Therefore, we have 4 possibilities: $|Z(G)| = \{1,p,p^2,p^3\}$

It can't be $p^3$ since that implies that $G$ is abelian, contradicting the initial assumption. There's another theorem that states that if $|G|$ is a power of a prime number, then the center of $G$ contains nonidentity elements, so it can't be $1$.

This leaves us with $p$ or $p^2$.

If the size is $p^2$, we can go back to the class equation and obtain $$p^3 = p^2 + \sum\frac{p^3}{|\mathrm{Cent}(g)|}$$ $$p\bigg( 1 - \sum\frac{p}{\mathrm{Cent}(g)}\bigg) = 1$$ $$\sum\frac{p}{\mathrm{Cent}(g)} = \frac{p-1}{p}$$

This number has to be an integer greater than or equal to $0$, and the only option is then $p=1$, but we already discarded that option, and $|Z(G)| \neq p^2$

Therefore, $|Z(G)| = p$ if $G$ is nonabelian.

If it is abelian then $G \setminus Z(G) = \emptyset$ so the sum is over no elements giving $0$, and $Z(G)=G \leq G$.

  • $\begingroup$ This looks correct to me, although I wouldn't phrase it as "Case 1": it suffices to prove that if $G$ is not abelian, then $\#Z(G) = p$ (that is logically equivalent to proving the given "or" statement), and so you may simply assume that $G$ is not abelian. [Another possibility to rule out $\#Z(G) = p^2$: use the fact that if $H$ is abelian and $G/H$ is cyclic, then $G$ is abelian.] $\endgroup$ Apr 13, 2018 at 7:55
  • $\begingroup$ @GregMartin if you understand that $\#Z(G)=|Z(G)|$, I suggest using the latter notation as perhaps the OP may not know the former. $\endgroup$
    – Mr Pie
    Apr 13, 2018 at 13:11
  • $\begingroup$ How did you get $$\sum\frac{p}{\operatorname{Cent}(g)} = \frac{p-1}{p}?$$If you start with the equation two lines above that and divide both sides by $p^2$, you get $$\sum\frac{p}{\operatorname{Cent}(g)} = p-1,$$ so where did you get the $p$ in the denominator? $\endgroup$
    – user169852
    Apr 13, 2018 at 13:22
  • $\begingroup$ $$p \bigg( 1- \sum(p / |\mathrm{Cent}(g)|) \bigg)$$ $$\sum p / | \mathrm{Cent}(g)| = 1 - 1/p = (p-1)/p$$ $\endgroup$
    – The Bosco
    Apr 13, 2018 at 17:30

2 Answers 2


Suppose that $G$ is non-abelian.

We know that $Z(G)\leq G$.

By Lagrange's Theorem $|Z(G)|$ must divide $|G|$.

Since $|G|=p^{3}$ the only possibilities are $1, p, p^{2}, p^{3}$.

$|Z(G)|\neq p^{3}$ because otherwise we will have $Z(G)=G$ but $G$ is non-abelian.

$|Z(G)|\neq p^{2}$ also because otherwise we will have the order of the factor group by the center as $|G/Z(G)|=|G|/|Z(G)|= p^{3}/p^{2} = p$. Therefore:

$|G/Z(G)|=p \implies G/Z(G)$ is cyclic $\implies G$ is abelian. But $G$ is non-abelian.

Now $|Z(G)|\neq 1$ also because $G$ is a $p-group$ and $p-groups$ have non-trivial center.

So, it must be that $|Z(G)|=p$.


The class equation you wrote is incorrect. You have to choose one element from each conjugacy class; if we denote by $g_1,\dots,g_k$ the ones chosen from the non singletons, we have $$ |G|=|Z(G)|+\sum_{i=1}^k\frac{|G|}{\lvert\operatorname{Cent}(g_i)\rvert} $$ This shows $|Z(G)|>1$, because $\operatorname{Cent}(g_i)\subsetneq G$, so the summation gives a multiple of $p$. Hence also $|Z(G)|$ must be a multiple of $p$. (Note that this just requires $|G|$ is a power of $p$.)

Thus only three cases are possible: $|Z(G)|=p$, $|Z(G)|=p^2$ or $|Z(G)|=p^3$. We need to exclude the second case.

Suppose $|Z(G)|=p^2$. Then $G/Z(G)$ has order $p$, so it's cyclic. If $gZ(G)$ is a generator, then for every $x\in G$ we have $xZ(G)=g^mZ(G)$, but this easily implies $G$ is abelian: contradiction.

I'm not sure your argument with the class equation is good.

  • $\begingroup$ For the class equation I copied straight from the book $\endgroup$
    – The Bosco
    Apr 13, 2018 at 17:33
  • $\begingroup$ @TheBosco If $|Z(G)|=p^2$, then $\operatorname{Cent}(g)>p^2$, so the summation need not be an integer. Indeed, it is $(p-1)/p$ which is definitely not an integer. Much more work is needed, in my opinion. $\endgroup$
    – egreg
    Apr 13, 2018 at 17:44
  • $\begingroup$ It is only an integer when p = 1, giving 0. It does have to be an integer isn't it? Since it is a finite group and both $G$ and $Z(G)$ have to have an integer amount of elements in them, therefore the sum has to be an integer as well. Also, you are right, the sum is on the conjugacy classes. That's what I meant but I wrote it wrong. That does not change the argument I wrote though $\endgroup$
    – The Bosco
    Apr 13, 2018 at 19:36
  • $\begingroup$ @TheBosco Why should the sum of $1/|\mathrm{Cent}(g_i)$ be an integer? $\endgroup$
    – egreg
    Apr 13, 2018 at 19:40
  • 1
    $\begingroup$ @TheBosco Yes, but you’re dividing it by $p^3$ $\endgroup$
    – egreg
    Apr 13, 2018 at 19:42

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