# p-group

• Dec 2nd 2011, 02:17 PM
dwsmith
p-group
Prove that if G is a finite p-group, then its center is non-trivial.
Use this fact to prove if $\displaystyle |G|=p^2$, then G is abelian.

For the first part, I have:
So $\displaystyle |G|=p^{\alpha}m$. By the class equation, $\displaystyle |G|=|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|$.

Since $\displaystyle p| \ |G|$, $\displaystyle p|\left(|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|\right)$

So $\displaystyle p| \ |Z(G)|$ and $\displaystyle p| \ \sum_{i=0}^r|G:C_G(g_i)|$.

Since p is prime, $\displaystyle |Z(G)|\neq 1$. Therefore, $\displaystyle Z(G)\neq\{e\}$

So how do I use that to prove the next part?
• Dec 2nd 2011, 06:03 PM
Drexel28
Re: p-group
Quote:

Originally Posted by dwsmith
Prove that if G is a finite p-group, then its center is non-trivial.
Use this fact to prove if $\displaystyle |G|=p^2$, then G is abelian.

For the first part, I have:
So $\displaystyle |G|=p^{\alpha}m$. By the class equation, $\displaystyle |G|=|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|$.

Since $\displaystyle p| \ |G|$, $\displaystyle p|\left(|Z(G)|+\sum_{i=0}^r|G:C_G(g_i)|\right)$

So $\displaystyle p| \ |Z(G)|$ and $\displaystyle p| \ \sum_{i=0}^r|G:C_G(g_i)|$.

Since p is prime, $\displaystyle |Z(G)|\neq 1$. Therefore, $\displaystyle Z(G)\neq\{e\}$

So how do I use that to prove the next part?

The first part is fine, but there is a principle you should remember. If $\displaystyle P$ is a finite $\displaystyle p$-group acting on a finite set $\displaystyle X$, and if $\displaystyle X^P$ denotes the fix set of the action (i.e. those elements that are fixed by every then $\displaystyle |X|\equiv |X^P|\text{ mod }p$. So, if, in your example, you let $\displaystyle X=G$ act on itself by conjugation you get that $\displaystyle |X^G|\equiv |G|=p^2\equiv 0\text{ mod }p$ but evidently $\displaystyle X^G=Z(G)$ and so the result follows. But, as I said, your solution is fine. This technique just generalizes and is nice to know.

For the second part you need to be aware of the common fact that if $\displaystyle N\leqslant Z(G)\leqslant G$ and $\displaystyle G/N$ is cyclic then $\displaystyle G$ is abelian (exercise). So, you know that $\displaystyle |Z(G)|=p,p^2$, suppose that $\displaystyle |Z(G)|=p$ then $\displaystyle |G/Z(G)|=p$ and so $\displaystyle G/Z(G)$ is cyclic, and so by the previous sentence $\displaystyle G$ is abelian, which contradicts that $\displaystyle |Z(G)|=p$. Thus, $\displaystyle |Z(G)|=p^2$ and so $\displaystyle G=Z(G)$ as desired.
• Dec 2nd 2011, 06:25 PM
dwsmith
Re: p-group
Why is |Z(G)|=p a contradiction? We want G to be abelian.
• Dec 2nd 2011, 06:33 PM
Drexel28
Re: p-group
Quote:

Originally Posted by dwsmith
Why is |Z(G)|=p a contradiction? We want G to be abelian.

It's a weird argument. You can just conclude by saying that $\displaystyle G$ is abelian, but it feels wrong. The contradiction is that if $\displaystyle G$ is abelian than $\displaystyle |Z(G)|=p^$.
• Dec 2nd 2011, 10:56 PM
Deveno
Re: p-group
i find it instructive to see why if N ⊆ Z(G), and G/N is cyclic, G is abelian.

let xN be the (a) generator of G/N. so every element of G can be written as: $\displaystyle x^kz$, where z is in Z(G) (because N is contained within Z(G)).

so for any g,h in G: $\displaystyle gh = (x^kz_1)(x^mz_2) = x^{k+m}z_1z_2$ (since $\displaystyle z_1$ is in Z(G))

$\displaystyle =(x^m)(x^k)z_2z_1 = (x^mz_2)(x^kz_1) = hg$ (since $\displaystyle z_2$ is in Z(G)), so G is abelian.

(the contradiction is: assuming |Z(G)| = p, and Z(G) ≠ G, leads to Z(G) = G. therefore, |Z(G)| ≠ p).