Let a,b,c be elements in G Prove the following.

Ord(a)=1 iff a=e.

Attempt: Could i just use definition of order of a group ? So if Ord(a)=1 then the size of G is a?

Results 1 to 3 of 3

- March 5th 2012, 08:11 PM #1

- Joined
- Sep 2010
- Posts
- 151
- Thanks
- 1

- March 5th 2012, 09:20 PM #2

- March 6th 2012, 01:31 PM #3

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,392
- Thanks
- 759

## Re: If Ord(a)=1 iff a=e

if o(a) = 1, this means that . so that's the "if" part.

suppose that a = e. then surely . so 1 is a positive integer k such that . since there is no positive integer smaller than 1, 1 is clearly the least positive integer with this property: that is, o(a) = 1.

to use this definition: o(a) = |<a>|, note that if |<a>| = 1, then <a> has just one element. this means that <a> contains the identity of G, since every subgroup of G contains the identity, and since a is the only element of <a>, a = e (if part). if a = e, then <a> = , and since for all integers k, <a> = <e> has just one element (namely, a = e), so |<a>| = |<e>| = 1 (only if part).