# Thread: Question about order of center

1. ## Question about order of center

Let the order of G be pq where p and q are primes (not necessarily distinct). Show that the center has order 1 or pq. Lets call the center of G Z(G)

Attempt:

By lagrange we know the order of Z(G) to be 1, p, q or pq. Now I must show it cannot be p or q.

I found a nifty couple of theorems in my algebra book that says: If G/Z(G) is cyclic then G is abelian. Using the contrapositive and lagrange we can show that a non abelian group of order pq must have a trivial center.

I don't know how to prove that second statement but it seems like that would be half of my proof. Then if I can show that an Abelian group with order pq must have a center of order pq i would be done.

Can you guys help, give some hints, or perhaps other ways you would do this? Thanks!!

2. If a group is abelian, then it is equal to its center by definition (the center of a group is the set of elements that commute with everything - if G is abelian, then every element commutes with everything).

Here is a proof that G/Z(G) cyclic implies G is abelian

Assume that G/Z(G) = <X> for some X in G/Z(G). So, X = xZ(G) for some x in G. If g is in G, then gZ(G) = (xZ(G))k = xkZ(G) for some integer k. So, g = xkz for some z in Z(G). If h is another element of G, then similarly we can write h = xry for some y in Z(G). Thus, gh = xkzxry = xkxrzy = xrxkyz = xryxkz = hg. Thus G is abelian.

3. Very good point! hahha gosh I should not do math when I am sick.

Could you help me prove that if G is non abelian then the center is trivial?

4. You should be able to show that if G is nonabelian, then G has an element of order p and an element of order q. If these two elements commute, then G is abelian. So p and q do not commute. Since p and q are prime, x^k is a generator for any k<p, and similarly for y. Thus, once again x^k commuting with y^m will imply G is abelian.

I haven't written down the details, but I think that this argument should work.

5. Originally Posted by mulaosmanovicben
Very good point! hahha gosh I should not do math when I am sick.

Could you help me prove that if G is non abelian then the center is trivial?

Otherwise the center has order p or q and then the quotient group $G/Z(G)$ has order q or p and is thus cyclic...

Tonio