Proving (ab)^n=e: A Group Theory Question | Homework Statement

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Homework Help Overview

The problem involves group theory, specifically proving that if \((ab)^n = e\) for elements \(a\) and \(b\) in a group, then \((ba)^n = e\) as well. The context suggests a focus on non-abelian groups.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the equation \((ab)^2 = e\) and explore how to manipulate the expression to derive \((ba)^n = e\). There are attempts to generalize the reasoning from specific cases to a broader proof.

Discussion Status

Some participants have offered insights into manipulating the expressions and using properties like associativity and cancellation. There is an acknowledgment of the challenge in generalizing the argument for all \(n\), with suggestions of using induction as a potential approach.

Contextual Notes

There is a recognition that the problem is set within the framework of non-abelian groups, which may influence the reasoning and approaches discussed.

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Homework Statement



If a and b are in a group, show that if (ab)^n=e then (ba)^n=e.

Homework Equations





The Attempt at a Solution



I'm not sure how one would prove this. The question is obviously for non-abelian groups.
 
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If (ab)^2=e then (ab)(ab)=e. So b(ab)(ab)a=bea=ba. Now use associativity and 'cancellation'. Do you see how to do the same trick for (ab)^n?
 
I'm not sure how this helps us show that (ba)^2=e? When generalizing to (ab)^n I see we'll get a similar result but I'm not sure how this shows that (ba)^n=e.
 
b(ab)(ab)a=ba, yes? That's the same as (ba)(ba)(ba)=(ba). Do you see it now?
 
abab=e , so ab=b-1a-1

ababab=e , so baba=a-1b-1

Play around with these until you can figure out one, then , if you don't have a general
argument for all n, maybe induction on n will help.
 
Oh, ok. Now I understand the argument. Thank you.
 

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