Prove that a cyclic group is abelian?

Answer 1

As mathematician would say homework! Here is a hint though, if the group is cyclic then there is an element a such that any b in the group is of the form a^m for some m. Now think about what you know if you multiply powers of a single element.

Answer 2

If G is cyclic, then at least one of its elements generates the whole group, that is, G has an element g (the generator) such that for any element a of G, there is an integer n such that a = g^n.
So let a and b be any two elements of G. Then there are integers n and m such that a = g^n and b = g^m, and
ab = (g^n)(g^m) = g^(n + m) = g^(m + n) = (g^m)(g^n) = ba.
Thus G is abelian.

Answer 3

recall what it means for a group to be cyclic
it means that the group can be generated by a single element

let our cyclic group be G = [a] (a is then referred to as a generator of the group)

then G is the group of all a^k with k an integer
..., a^-2, a^-1, e, a^1, a^2, ....

so let x, y be elements of G
then x = a^m and y = a^n with m and n integers

so determine xy and yx and show that xy=yx

[and your proof would be done since x,y are any arbitrary elements from G]