p-group

**dwsmith** · December 2nd 2011, 02:17 PM

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

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

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

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

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

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

**Drexel28** · December 2nd 2011, 06:03 PM

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 $|G|=p^2$ , then G is abelian.

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

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

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

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

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 $P$ is a finite $p$ -group acting on a finite set $X$ , and if $X^P$ denotes the fix set of the action (i.e. those elements that are fixed by every then $|X|\equiv |X^P|\text{ mod }p$ . So, if, in your example, you let $X=G$ act on itself by conjugation you get that $|X^G|\equiv |G|=p^2\equiv 0\text{ mod }p$ but evidently $X^G=Z(G)$ 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 $N\leqslant Z(G)\leqslant G$ and $G/N$ is cyclic then $G$ is abelian (exercise). So, you know that $|Z(G)|=p,p^2$ , suppose that $|Z(G)|=p$ then $|G/Z(G)|=p$ and so $G/Z(G)$ is cyclic, and so by the previous sentence $G$ is abelian, which contradicts that $|Z(G)|=p$ . Thus, $|Z(G)|=p^2$ and so $G=Z(G)$ as desired.

**dwsmith** · December 2nd 2011, 06:25 PM

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

**Drexel28** · December 2nd 2011, 06:33 PM

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 $G$ is abelian, but it feels wrong. The contradiction is that if $G$ is abelian than $|Z(G)|=p^$ .

**Deveno** · December 2nd 2011, 10:56 PM

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: $x^kz$ , where z is in Z(G) (because N is contained within Z(G)).

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

$=(x^m)(x^k)z_2z_1 = (x^mz_2)(x^kz_1) = hg$ (since $z_2$ 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).

Thread: p-group

LinkBack

Thread Tools

Display

p-group

Re: p-group

Re: p-group

Re: p-group

Re: p-group

Similar Math Help Forum Discussions

Order of Group. Direct Product of Cyclic Group

Group Theory-Show no infinite soluble group has a composition series

Quick questions on Group Theory - Cosets / Normal Group

Group, Cosets, Normal Group help

Group Theory Question, Dihedral Group