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Proofs on the Order of Group Elements


Date: 10/19/98 at 20:40:04
From: Jennifer V.D. Kopek
Subject: Order of group elements

Hi, 

I need help getting started on these three problems. 

Let a be any element of finite order of a group G. Prove:

1. If a^p = e where p is a prime number, then a has order p. 
   (a doesn't equal e)

2. The order of a^k is a divisor of the order of a.
   I got this far: let ord(a) = n. 
   Then (a^k)^n = a^(nk) = (a^n)^k = e^k = e.
    
3. If ord(a) = km, then ord(a^k) =  m.

Any help would be great.
Thanks, JVDK.


Date: 10/20/98 at 19:51:29
From: Doctor Tom
Subject: Re: Order of group elements

Hi Jennifer,

1. Clearly the order is <= p, since the order of a group element is 
the smallest k such that a^k = e. Since a^p = e, either p is the 
order, or there is some k less than p such that a^k = e.

Since p is prime, if there is such a k, then k doesn't divide p, so 
p = nk + m, and m is less than k.

e = a^p = a^(nk+m) = a^(nk)a^m = e^na^m = a^m, so a^m = e, so k wasn't 
the order after all. Thus p must be the order of a.

2. If the order of a^k does not divide the order of a, suppose the 
order of a^k = m. a^(km) = e and m is the smallest number such that 
this is true.

If the order of a is n, then a^n = e. If m doesn't divide n then
n = qm + r, with r < m. Note:

   e = a^(nk) = (a^k)^(qm+r) = a^(qmk)a^r = ea^r = a^r

and r < m, so our assumption that m is the smallest number such that 
a^m = e is false. Thus the order of a^k divides the order of a.

3. If ord(a) = km, then clearly a^km = e, so the order of a^k is less 
than or equal to m. If it's less, say q, then (a^k)^q = a^kq = e, and 
kq < km, so then the order of a is also less than km, since a^kq = e.

- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Modern Algebra

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