Proving the Nontriviality of the Center of a Finite p-Group

[Locoism]

Homework Statement

Let G be a finite p-group, where p is a positive prime. Show that G has nontrivial center. In other words Z(G) \neq {e}.

The Attempt at a Solution

So the centre is pretty much the "abelian subgroup" of G, or all the elements that commute with every other element. Now I remember that if G has prime order, then it is abelian, but I can't find the proof, although I'm not sure if that was a "if and only if" statement or if it was one sided... So if I can prove that, then Z(G) is just equal to G?

 

[Deveno]

a p-group is not of prime order, it's of prime power order (like 8, or 27, or 3,125).

the trick is to use the class equation:

|G| = |Z(G)| + \sum_{a \not \in Z(G)} [G:N(a)]

where N(a) is the normalizer of a non-central element.

note that since N(a) is a subgroup of G for each a not in the center, it also has prime power order. why is it true that each N(a) is strictly smaller then G (hint: a is not in the center)?

use this to show that p divides each normalizer index in the sum. you should be able to show that p then also divides |Z(G)|.

of course, p divides 0, so you have to eliminate this possibility. why is this trivial (can the center be empty)?

 

[Locoism]

Oh ok so clearly N(a) is smaller than G because the identity is in the centre by definition, and p divides |G| so it must divide |Z(G)| (I'll work that out later) and Z(G) isn't empty since automatically it has e,
But I don't understand: G is a finite p-group means it is of prime power order and not of prime order. What is the difference? How is 8, 27, and 3,125 of prime power order?

 

[Deveno]

Oh ok so clearly N(a) is smaller than G because the identity is in the centre by definition, and p divides |G| so it must divide |Z(G)| (I'll work that out later) and Z(G) isn't empty since automatically it has e,
But I don't understand: G is a finite p-group means it is of prime power order and not of prime order. What is the difference? How is 8, 27, and 3,125 of prime power order?

no, the reason N(a) is smaller than G is NOT "because the identity is in the center". that IS, however, the reaon the center cannot be empty.

N(a) = {g in G: ga = ag}

think about this: why can't N(a) be all of G? what would that mean for a?

saying "you'll work out later" why p must divide |Z(G)| means you're missing the whole point of the proof. you're trying to show |Z(G)| is non-trivial. that's the same thing as saying p divides |Z(G)|...it's not a minor point of the proof...it's the HEART.

8 = 2^3 <--- 2 is a prime.
27 = 3^3 <--- 3 is a prime.
3,125 = 5^5 <--- 5 is a prime.

each of these numbers is a "power of a prime", in the first 2 cases p³, for the last one, p⁵.

a p-group is a group where |G| = p^k, for some positive integer k (which MIGHT be 1, but it might be some larger positive integer).

 

[Locoism]

Ah well then if N(a) is all of G, then a would be in Z(G). And in that case p clearly divides |G| (leave p^k-1) so p must divide Z(G) and N(a), so since Z(G) is not empty, it must be at least order p, which isn't 1.

Thank you!

One last small question, is it true that a group with prime order is abelian or that a p-group is abelian? Is it a statement that goes both way or is it only one directional? I remember reading that but I can't find it...