Show that an element of a group has order n if and only if it generates a cyclic group of order n?

An element ( g ) of a group ( G ) has order ( n ) if the smallest positive integer ( k ) such that ( g^k = e ) (the identity element) is ( n ). This means the powers of ( g ) generate the set ( { e, g, g^2, \ldots, g^{n-1} } ), which contains ( n ) distinct elements. Therefore, the cyclic group generated by ( g ), denoted ( \langle g \rangle ), has exactly ( n ) elements, thus it is a cyclic group of order ( n ). Conversely, if ( \langle g \rangle ) is a cyclic group of order ( n ), then ( g ) must also have order ( n ) since ( g^n = e ) is the first occurrence of the identity.