# P-group problem group theory

1. Jan 31, 2015

### MostlyHarmless

1. The problem statement, all variables and given/known data
Let G be a non-abelian group with order $p^3$, p prime. Then show that the order of the center must be p.

2. Relevant equations
Theorem in our book says that for any p-group the center is non-trivial and it's order is divisible by p.
Class eq.
$|G|= |Z(G)| + \sum{[G : C(x)]}$ Where, [G: C(x)] is the number of left cosets of C(x) in G. And $C(x) := (a\in G | axa^{-1}=x)$ is the centralizer. And we are summing over x by taking a single x in each of the non-trivial conjugacy class of G and looking at that centralizer.

3. The attempt at a solution
My thought was to do a counting argument, but it felt really weak.

So by the theorem in the book, since G is a p-group we have that Z(G) is non trivial and p | |Z(G)|, and obviously the order of Z(G) is strictly less that G since G is non-abelian. So we get that |Z(G)| must be either $p$ or $p^2$.

So, for contradiction, suppose that $|Z(G)|=p^2$. So we have $$p^3= p^2+ \sum{[G : C(x)]}$$ $$p^3-p^2=\sum{[G : C(x)]}$$
From here I apply Lagrange's theorem $[G:C(x)]=\frac{|G|}{|C(x)|}$ and reason that since any centralizer has at the very least, the center of the group, $$\frac{|G|}{|C(x)|}<\frac{p^3}{p^2}=p$$. I further reason that there are $p^3-p^2$ non-central elements, and since a non-trivial conjugacy class has at least 2 elements in it. That there are at most $\frac{p^3-p^2}{2}$ non-trivial conjugacy classes. So the sum$$\sum{[G : C(x)]}<\frac{p^3-p^2}{2}p$$ Giving the inequality:$$p^3-p^2<\frac{p^3-p^2}{2}p$$. Which fails when p=2. Thus a contradiction.

The reason this feels weak, is because it doesnt fail for p=3, or p=5, or any other primes I tried.

A classmate mentioned an argument using the fact that a group mod its center is abelian, but I'm not sure that is true in general. I couldn't find it anywhere.

2. Jan 31, 2015

### Dick

Think about the factor group $G/Z(G)$ when $|Z(G)|=p^2$. A group mod its center isn't always abelian. But this one is. In fact, it's cyclic. What can you do with that?

Last edited: Jan 31, 2015
3. Jan 31, 2015

### MostlyHarmless

Well, if G/Z(G) is cyclic, then its trivial, meaning that G is abelian, which is a contradiction.

How do we know that G/Z(G) is cyclic in this case?

4. Jan 31, 2015

### Dick

What's the order of $G/Z(G)$?

5. Feb 10, 2015

### MostlyHarmless

Sorry for the delay in response. In this case, G/Z(G) had order p-prime. Thus, it is cyclic.

6. Feb 10, 2015

Of course.