Prove that if G is a finite p-group, then its center is non-trivial.
Use this fact to prove if , then G is abelian.
For the first part, I have:
So . By the class equation, .
So and .
Since p is prime, . Therefore,
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 is a finite -group acting on a finite set , and if denotes the fix set of the action (i.e. those elements that are fixed by every then . So, if, in your example, you let act on itself by conjugation you get that but evidently and so the result follows. But, as I said, your solution is fine. This technique just generalizes and is nice to know.
Originally Posted by dwsmith
For the second part you need to be aware of the common fact that if and is cyclic then is abelian (exercise). So, you know that , suppose that then and so is cyclic, and so by the previous sentence is abelian, which contradicts that . Thus, and so as desired.
Why is |Z(G)|=p a contradiction? We want G to be abelian.
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: , where z is in Z(G) (because N is contained within Z(G)).
so for any g,h in G: (since is in Z(G))
(since 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).