Question

Prove that, if $\mid G \mid P^{n}$, where p is a prime then G has non-trivial center.

Answer 1

hii baby

Prove that, if $\mid G \mid P^{n}$, where p is a prime then G has non-trivial center.

I'm trying to understand the proof of a Burnside theorem (as stated in Beachy's Abstract Algebra p. 328): Let p be prime number. The center of any p-group is nontrivial.

Now, In the proof they say that if we let G be a p-group, then in the class equation

|G|=|Z(G)|+∑[G:C(x)]

for all x that is not in the center and represent a conjugacy class, we see that every term in ∑[G:C(x)] is divisible by p since x∉Z(G)⟹[G:C(x)]>1. This last statement is what I do not understand, how do we know that p∣[G:C(x)] for any conjugacy class?

I know that the elements in the conjugacy class of x is in bijection with the cosets of C(x), i.e. [G:C(x)], but how can we be certain that the number of elements in a conjugacy class of x/cosets of the centralizer of x is divisible by p?

Best regards.