- #1

MostlyHarmless

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## Homework Statement

Let G be a non-abelian group with order ##p^3##, p prime. Then show that the order of the center must be p.

## Homework 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.

## 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.