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?
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).