Proving that 'a' can be written as the nth power

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Homework Help Overview

The discussion revolves around a problem in group theory, specifically regarding the expression of an element 'a' as the nth power of another element in a group, given that 'n' is relatively prime to the order 'm' of 'a'.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the greatest common divisor and the ability to express 'a' as an nth power. There are attempts to manipulate equations involving linear combinations of 'm' and 'n' to derive the desired expression.

Discussion Status

Some participants provide insights and corrections to each other's reasoning, particularly regarding the manipulation of powers and the implications of the gcd condition. There is a progression towards a clearer understanding of how to express 'a' as an nth power, although not all steps are fully resolved.

Contextual Notes

Participants express uncertainty about algebraic manipulations and the clarity of their reasoning, indicating a struggle with the material. There is a mention of a preference for different types of homework, suggesting varying levels of comfort with the subject matter.

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Homework Statement



Let a belong to a group and |a| = m. If n is relatively prime to m, show that
a can be written as the nth power of some element in the group.

Homework Equations



We know :
|a| = m
gcd(m,n) = 1

We want to show :
bn = a for some b in the group.

The Attempt at a Solution



Okay, I didn't really know where to start with this one, but I'll give it a try.

We know the gcd can be written as a linear combination, that is :

gcd(m,n) = 1 = ms + nt for some integers s and t.

Now :
1 = ms + nt
a1 = ams + nt
a = ams ant
a = easant
a = asant

Here's where I get stuck. Any pointers?
 
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Zondrina said:

Homework Statement



Let a belong to a group and |a| = m. If n is relatively prime to m, show that
a can be written as the nth power of some element in the group.

Homework Equations



We know :
|a| = m
gcd(m,n) = 1

We want to show :
bn = a for some b in the group.

The Attempt at a Solution



Okay, I didn't really know where to start with this one, but I'll give it a try.

We know the gcd can be written as a linear combination, that is :

gcd(m,n) = 1 = ms + nt for some integers s and t.

Now :
1 = ms + nt
a1 = ams + nt
a = ams ant
a = easant
I'm not sure what you did to get the last line. a^{ms} = (a^m)^s = e^s = e, so the last line should be a = a^{nt}.
 
jbunniii said:
I'm not sure what you did to get the last line. a^{ms} = (a^m)^s = e^s = e, so the last line should be a = a^{nt}.

Ohhh whoops. I got confused by all the tags lol.

So really I have :

1 = ms + nt
a1 = ams + nt
a = ams ant
a = (am)sant
a = esant
a = ant

Hence a can be written as the nth power of some element in the group and we are done.
 
Zondrina said:
Ohhh whoops. I got confused by all the tags lol.

So really I have :

1 = ms + nt
a1 = ams + nt
a = ams ant
a = (am)sant
a = esant
a = ant

Hence a can be written as the nth power of some element in the group and we are done.


Right, specifically a = (a^t)^n.
 
jbunniii said:
Right, specifically a = (a^t)^n.

Yeah I wrote that down for my final solution after, I was just highlighting that your hint let me get there nice and quickly.

I wish I had analysis homework rather than this. I feel lost with algebra all the time.
 

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