If Ord(a)=1 iff a=e

**Aquameatwad** · March 5th 2012, 08:11 PM

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?

**beebe** · March 5th 2012, 09:20 PM

If a,b,c are distinct elements, then the order of G is 3 or greater. Use the definition of order for group elements. That is, $o(x)=n$ for the smallest positive integer n which satisfies $x^n=e$ .

**Deveno** · March 6th 2012, 01:31 PM

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

suppose that a = e. then surely $a^1 = a = e$ . so 1 is a positive integer k such that $a^k = e$ . 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> = $\{e^k : k \in \mathbb{Z}\}$ , and since $e^k = e$ for all integers k, <a> = <e> has just one element (namely, a = e), so |<a>| = |<e>| = 1 (only if part).

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