No problem! Keep working at it and it will become second nature.

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Homework Help Overview

The discussion revolves around the order of an element \( a^k \) in a group \( G \), where \( a \) is an element of order \( m \). Participants explore the relationship between the orders of \( a \) and \( a^k \), particularly focusing on the least common multiple of \( k \) and \( m \).

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the order of \( a^k \) and the least common multiple of \( k \) and \( m \). There is a focus on clarifying the notation used, particularly regarding the correct terms for the least common multiple. Some participants also explore the implications of \( \text{lcm}(k, m) = km \) on the greatest common divisor of \( k \) and \( m \).

Discussion Status

The discussion is active, with participants confirming each other's understanding and clarifying points of confusion. There is an acknowledgment of a typo in the notation, and participants are engaging with the implications of their findings regarding the orders of elements in the group.

Contextual Notes

Participants note that the original poster is relatively new to group theory, which may influence their understanding and interpretation of the concepts discussed.

radou
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Homework Statement



Let G be a group, and a an element of G of order m. What is the order of a^k?

The Attempt at a Solution



Well, first of all, if (a^k)^p = e, for some p, we have kp = mq, for some q. Now, for some k, the order of a^k is the least such p. Hence, it would make to sense to consider the least common multiple of k and m, lcm(m, n). The lcm(m, n)/k is the order of a^k. This is a bit informal perhaps, but it seems clear to me. Is this correct?
 
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Further on, it would follow from this that the powers of a which have the same order as a are those such that the least common multiple of k and m equals km.
 
Seems clear to me too. But you meant lcm(m,k), right? There's no 'n' in the problem.
 
Dick said:
Seems clear to me too. But you meant lcm(m,k), right? There's no 'n' in the problem.

Oh, sorry, it was a typo - yes, I meant lcm(m, k). Thanks.
 
radou said:
Further on, it would follow from this that the powers of a which have the same order as a are those such that the least common multiple of k and m equals km.

And if lcm(k,m)=km that tells you something about gcd(k,m).
 
Dick said:
And if lcm(k,m)=km that tells you something about gcd(k,m).

Well, integers k, m such that lcm(k, m) = km tells us that k and m are relatively prime, hence gcd(k, m) = 1... I don't see a fact which follows from this right away.
 
radou said:
Well, integers k, m such that lcm(k, m) = km tells us that k and m are relatively prime, hence gcd(k, m) = 1... I don't see a fact which follows from this right away.

Nothing you haven't already said. But the phrasing most people would use is that a^k generates the whole group if k and m are relatively prime.
 
Dick said:
Nothing you haven't already said. But the phrasing most people would use is that a^k generates the whole group if k and m are relatively prime.

Yes, I'm aware of this fact - I proved it in another exercise. I'm still a bit new to group theory, so I don't realize some things right away yet. :smile:
 

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