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ROBBYM

Member

Registered: 2008-04-24

Posts: 10

algebra

a and b are elements of group G. show that ab and ba have the same order. to complete this question iam suppose to find a reationship between (ab)^n and (ba)^n using the general associative law

I'm a little confused how to do this question as i would have used the comutative law which must be wrong because the question is worth a lot of marks please help me i have an exam on algebra soon and its quite worrying i can't answer this question

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: algebra

e.g.   n=5
(ab)(ab)(ab)(ab)(ab)
use associate law.
a(ba)(ba)(ba)(ba)b
That's all I can think of.

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igloo myrtilles fourmis

cushydom

Member

Registered: 2008-04-23

Posts: 10

Re: algebra

suppose that ab and ba have different orders the one of then must have an order greater then the other. let the order of ab be greater than ba. let n be the order of ab. so (ab)^n=1.
so (ab)(ab)......(ab)(ab)=1 (with n lots of ab)
a(ba)(ba)......(ba)(ba)b=1   since the order of ba is less than that of ab some combination of these ba must equal one. this gives:
(ab)(ab)......(ab)(ab)=1 (with less than n lots of ab)
this gives the order of ab less than n. contradiction. therfore the order of ab = the order of ba

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: algebra

Nice proof!
I just want to jump in late and mention that commutivity doesn't hold for groups in general, so ROBBY's initial thought wouldn't work.

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Why did the vector cross the road?
It wanted to be normal.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: algebra

Direct method:  Let o(ab) = n

So note that

Now:

This proves that o(ba) | o(ab).  Do the steps again starting with o(ba) = m to get o(ab) | o(ba) and conclude o(ab) = o(ba).

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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..."