Associated Topics || Dr. Math Home || Search Dr. Math

### 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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search