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

In summary, the centre of G is a nontrivial element that is composed of all the elements that commute with every other element.
  • #1
Locoism
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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) [itex]\neq[/itex] {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?
 
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  • #2
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:

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

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)?
 
  • #3
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?
 
  • #4
Locoism said:
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| = pk, for some positive integer k (which MIGHT be 1, but it might be some larger positive integer).
 
  • #5
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 pk-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...
 

1. What is the "center" of a group of order p?

The center of a group of order p, denoted as Z(G), is the set of elements in the group that commute with every other element in the group. In other words, for any element a in Z(G) and any element b in the group G, ab = ba.

2. How is the center of a group related to its order p?

The order of the center, denoted as |Z(G)|, is always a divisor of the group's order p. This means that the number of elements in the center must be a factor of the total number of elements in the group.

3. Is the center of a group of order p always non-empty?

Yes, the center of any non-trivial group (a group with more than one element) is always non-empty. This is because the identity element, which is always part of a group, commutes with all other elements and is therefore always in the center.

4. Can the center of a group of order p be larger than its subgroup of order p?

Yes, it is possible for the center to be larger than its subgroup of order p. This can happen when there are multiple elements in the center that commute with each other, creating a larger subgroup within the center itself.

5. How is the center of a group of order p related to its normal subgroups?

The center of a group of order p is always a normal subgroup of the group. This means that the center is closed under conjugation, which is the process of multiplying an element by another element and its inverse. Additionally, the center is the largest normal subgroup of the group.

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