Element of a group composed a times = e -> a = e

  • #1

Homework Statement


EDIT: Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.


Homework Equations





The Attempt at a Solution


I know that if GCD(m,n) =1 then if a^m = e, then o(a) | m, but by Lagrange, o(a) | n -> 1 = GCD(m,n) >= o(a) -> o(a) =1 -> a=e.

However, if a^m=e -> a=e, I thought the following: suppose d=GCD(m,n). Then, d= bm + cn (b,c integers) suppose a^m = e -> a^bm = e -> a^(bm + cn) = e -> a^d =e, then I thought so now we need a^d =e -> a=e, so o(a) can't divide d, but o(a) divides m and n, since a^m =e, whence o(a) | bm +cn -> o(a)|d

(but if o(a)|m, then a^m =e doesn't imply a= e) so I don't know if it is right to say o(a)|m, so I am stuck here and any help would be appreciated.

Thanks

EDIT: Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.
 
Last edited:

Answers and Replies

  • #2
mathwonk
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no elements of finite order except e. for example the group of non zero real numbers, under multiplication.
 
  • #3
Deveno
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6

Homework Statement



I was wondering about a necessary and sufficient condition for a^m = e -> a=e.

Homework Equations





The Attempt at a Solution


I know that if GCD(m,n) =1 then if a^m = e, then o(a) | m, but by Lagrange, o(a) | n -> 1 = GCD(m,n) >= o(a) -> o(a) =1 -> a=e.

However, if a^m=e -> a=e, I thought the following: suppose d=GCD(m,n). Then, d= bm + cn (b,c integers) suppose a^m = e -> a^bm = e -> a^(bm + cn) = e -> a^d =e, then I thought so now we need a^d =e -> a=e, so o(a) can't divide d, but o(a) divides m and n, since a^m =e, whence o(a) | bm +cn -> o(a)|d

(but if o(a)|m, then a^m =e doesn't imply a= e) so I don't know if it is right to say o(a)|m, so I am stuck here and any help would be appreciated.

Thanks
necessary and sufficient condition on...what? a? m? the group G?
 
  • #4
mathwonk
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good point. my answer was for m>1. if m = 1, we could say any group.
 
  • #5
Thanks so much for your responses, I forgot to say, sorry, a necessary and sufficient condition on m and the order of the abelian group n. Sorry for leaving out so much, just spent so much time on it, I didn't even ralize I left the details out. Again:

Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.
 
  • #6
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By contraposition. Given [itex](m,n)\neq 1[/itex], try to find an element [itex]a\neq e[/itex] such that [itex]a^m=e[/itex].
 
  • #7
I've looked at the fact that GCD(m,n) is a linear combination of them, and said something like: Let d=(m,n) then d=bm + cn, b,c integers, then
a^(d)= a^(bm +cn) = a^bm -> o(a)|d-bm, but we already knew this by Lagrange.

a^(d-1)=/ e since d>1, but a^m*(d-1) = a^m*(bm +cn -1) = a^m*(bm-1), but I dont know how to show this is identity.


EDIT just thought d|m, d|n -> m=q'd, n=qd -> a^(n/d) = a^q, if I had some way to show o(a)|\ q, then I would have a^q =/ e, whence a^(q*m) = a^(n/(m/q`))*m = a^(n*q') = e, but I need to have that o(a) doesn't divide q, I know that if d>1, then n>q, but what if q=k*o(a) for some integer k??
 
  • #8
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I've looked at the fact that GCD(m,n) is a linear combination of them, and said something like: Let d=(m,n) then d=bm + cn, b,c integers, then
a^(d)= a^(bm +cn) = a^bm -> o(a)|d-bm, but we already knew this by Lagrange.

a^(d-1)=/ e since d>1, but a^m*(d-1) = a^m*(bm +cn -1) = a^m*(bm-1), but I dont know how to show this is identity.
Do you know Cauchy's theorem?? Or sylows theorems??? Look them up and try to use them in your reasoning.
 
  • #9
Do you know Cauchy's theorem?? Or sylows theorems??? Look them up and try to use them in your reasoning.
Thanks, but

I do actually know both, but Cauchy's thm. only applies to finite groups whose orders are divisible by a prime, and Sylow's only to a finite group of order p^n *m, where p doesn't divide m, so given n arbitrary, I don't think the proof should require these, (I wanted to do a general case) but does my edit on the last post loook right?
 
  • #10
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Thanks, but

I do actually know both, but Cauchy's thm. only applies to finite groups whose orders are divisible by a prime,
This is every group. (except the trivial group). Every number n>1 is divisible by a prime.

and Sylow's only to a finite group of order p^n *m, where p doesn't divide m, so given n arbitrary,
This is again every group. (except the trivial group). Every number can be written in those ways.

I don't think the proof should require these, (I wanted to do a general case) but does my edit on the last post loook right?
What's bothering me about it is that you never actually construct a. Given n and m you must FIND an [itex]a\neq a[/itex] such that [itex]a^m=e[/itex]. So, how would you make a?
 
  • #11
Thanks, yeah, by the thm. n>1 -> n is prime or product of primes, Cauchy's Thm. can be applied, but still I am not quite sure I understand what you mean by find. If we don't know what the group is, what I thought find meant was in terms of some arbitrary non identity element a??
 
  • #12
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Thanks, yeah, by the thm. n>1 -> n is prime or product of primes, Cauchy's Thm. can be applied, but still I am not quite sure I understand what you mean by find. If we don't know what the group is, what I thought find meant was in terms of some arbitrary non identity element a??
The element a is obviously not arbitrary. Given n and m, you must show that there EXISTS an [itex]a\neq e[/itex] such that [itex]a^m=e[/itex].

Indeed, the group is general, so you don't really have much options in constructing your a. Cauchy's theorem will be helpful as it states the existence of elements with certain properties...
 
  • #13
Thanks,
Ok, sorry for my silly mistakes,

By Cauchy, we know there exists an x in G with o(x) = p, where p is some prime factor of n. If p>d, p cannot divide d, so (p,d) =1, but d|n, p|n , (p,d) =1 -> dp|n -> x^(dp) = e. Still working, but is this the right track?
 
  • #14
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Thanks,
Ok, sorry for my silly mistakes,

By Cauchy, we know there exists an x in G with o(x) = p, where p is some prime factor of n. If p>d, p cannot divide d, so (p,d) =1, but d|n, p|n , (p,d) =1 -> dp|n -> x^(dp) = e. Still working, but is this the right track?
Yes, but which p did you take??
 
  • #15
As written, I guess I took an arbitrary prime divisor of n, and am checking if >d or <d, but (p,d) =1 -> x^d =/ e, but x^md = x^lp
d=am + bn = am + bp/q
x^d*m = x^am^2
 
  • #16
Am I supposed to take the least prime divisor exists by law of well-ordering on Z+, and then arrive at contradiction on minimality?
 
  • #17
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Remember that you assumed [itex]gcd(m,n)\neq 1[/itex] Do something with that.
 
  • #18
dp|n -> dp can't divide m, since if it did, it would be (m,n),
but x^d = x^(am+bn) = x^am=/e since (p,d) =1.
 
  • #19
d|n, so let n=q*d. Let x be the above element, then if p>q, p|/ q -> x^q =/e, but x^m*q = x^(q'*d) *q =x^(q'*n_ = e, a contradiction. So, p must divide q. But if p|q ....??
 
  • #20
If p|q, p<q, whence x^q =/ e, since p is the smallest integer which satisfies x^ y = e !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Is this right?
 
  • #21
Suppose a^m =e -> a=e, then suppose d>1 is s.t. d=(m,n). d>1 -> d is a prime or a product of primes. If d is a prime, since d|n, by Cauchy's Thm, there exists an element, x in G s.t. o(x) =d. But then since d|m, x=/e, but x^m =e, a contradiction. So, then d is a product of primes. But d|n -> there are elements of orders p1 and p2 (pi are prime divisors of d|n), but then by Cauchy's Thm, there are elements of order pi in the group. d|m , pi relatively prime, p1p2|m -> p1|m, p2|m, whence either of these elements are counter examples to the assumption. Therefore, we have a contradiction on d>1 and d=1. QED



Thanks to everyone's help.
 

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