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Group Theory Help...


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#1 Heretic Monkey

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Posted 24 October 2007 - 11:50 PM

Hey all. There was once a time where i graced these forums daily, offering my wealth of knowledge and aid to those that needed it...... well, i'm sure everyone else would say something different though.... lol. The truth is that i became heavily involved in lots of other things outside the internet (weird, right?), and just haven't had the time to cruise the forums here at good ol' BC. However, i come back to the intelligent people here in my time of need.....

I'm in MA 407 (abstract/modern algebra) at NCSU and need major help with a sample problem.... Namely:
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If |G| (the order of group G) = pq, where p and q are primes that are not necessarily distinct, prove that |Z(G)| (the order of Z(G), the center of G) = 1 or pq.

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I understand that in the broad sense, |Z(G)| could be 1, p, q, or pq, but p and q must be able to be "counted out" somehow. I just don't know the proof of how to eliminate p and q as possible solutions to the order.....

Again, any help would be appreciated. I think this is the best place to post the problem, since it really doesn't have all that much to do with electronics of any kind. Hopefully someone here has enough experience with group theory to help me out.... :thumbsup:

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#2 boopme

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Posted 25 October 2007 - 06:07 PM

Just wanted to say HELLO, HM. I can't answer yoiur question. Good to hear from you. By the way you've gotten a bit heavy in that avi... lol
How do I get help? Who is helping me?For the time will come when men will not put up with sound doctrine. Instead, to suit their own desires, they will gather around them a great number of teachers to say what their itching ears want to hear....Become a BleepingComputer fan: Facebook

#3 Heretic Monkey

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Posted 27 October 2007 - 04:42 PM

Just wanted to say HELLO, HM. I can't answer yoiur question. Good to hear from you. By the way you've gotten a bit heavy in that avi... lol

Yeah, well you know i can't stay away from a cold pawtucket patriot...

Anyway, i wasn't sure who would've been able to help me, it was just a long shot. I've been lurking in the shadows for a while, basically just searching for questions i've had that have already been answered, and i haven't posted in a while. Although i'm having another issue with my mouse, so i'm about to go post in the hardware section.... lol

#4 Albeda

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Posted 31 October 2007 - 02:44 PM

I've been working on this same math problem, and think I have it... see if you can poke holes in my answer:

If |G| = pq, where p and q are primes that are not necessarily distinct, prove that |Z(G)| = 1 or pq.

- We know Z(G) is a subgroup of G, and G is a finite group (since its order is finite). By Lagrange's theorem, then, Z(G) must be 1, p, q, or pq.
- There are two cases for G: G is Abelian, or G is non-Abelian.
- If G is Abelian, Z(G) = G, so |Z(G)| = |G| = pq.
- If G is non-Abelian, |Z(G)| = 1, p, or q.
- Assume |Z(G)| = p. By Lagrange, |G/Z(G)| = |G| / |Z(G)| = q, a prime. Groups of prime order are cyclic, so G/Z(G) is cyclic. By G/Z theorem, then, G is Abelian, a contradiction. Therefore, |Z(G)| is not p. Similarly, there is a contradiction for |Z(G)| = q. Therefore, if G is non-Abelian, |Z(G)| = 1. (This is alluded to in the Gallian text in the paragraphs following the proof of the G/Z theorem-- "it follows immediately from this statement and Lagrange's Theorem that a non-Abelian group of order pq, where p and q are primes, must have a trivial cener.")
- Therefore, |Z(G)| = 1 or pq.

Edited by Albeda, 31 October 2007 - 07:05 PM.


#5 cowsgonemadd3

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Posted 08 November 2007 - 12:30 AM

What is that stuff? I want to know so if/when I go to college I be sure and steer clear of over the top math.

PS welcome back and dont be a stranger...

Edited by cowsgonemadd3, 08 November 2007 - 12:30 AM.





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