Math Is Fun Forum

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Group order proof #2

Im stuck on this one:

And im stuck. I havent done anything basically, just laid out the background for the proof. Can someone help?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Group order proof #2

Maybe Ricky has a better proof than this. Ricky?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Group order proof #2

I'll leave it to you to judge whether or not this is "better".

As Jane did, assume a is not 1.  Let the order of x be n.  Then for all integers m, (x^m)^n = 1.  That is, a^n = 1, giving that k divides n.  Now k | n | pk and hence either n = k or n = pk.  If k = n, then 1 = x^n = x^k = a, but we assumed a is not 1.  Therefore, n = pk.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Group order proof #2

Well, it would seem that a large part of the proof has to do with number theory and divisibility rather than directly with group theory.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Group order proof #2

Of course once you find a property in a field of mathematics that involves divisibility, you're going to be doing some number theory.  That is unavoidable.

But there is nothing wrong with a number theoretic proof.  I tend to like proofs where fields branch over: algebra in analysis, field theory in number theory, etc.  Actually, one of my favorite proofs is a topological proof on the infinite amount of primes.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."