Showing a property of Abelian groups of order n

[Mr Davis 97]

Homework Statement

Let G be an abelian group of order n, and let k be an nonnegative integer. If k is relatively prime to n, show that the subgroup generated by a is equal to the subgroup generated by a^k

Homework Equations

The Attempt at a Solution

I'm not sure where to start. I know that we are equating two sets, so I think that I need to show that one is a subset of the other and vice versa, but I can't see where to use the fact that G is abelian and that k is relatively prime to n.

 

[andrewkirk]

I need to show that one is a subset of the other and vice versa

One of those directions is very easy, and doesn't need to use the 'relatively prime' property. Which one?

Write out the answer to that easy direction and you may find that it gives you ideas about how to complete the opposite direction, noting that you almost certainly will need to use the relatively prime property for that opposite direction. If you get stuck, hints may be forthcoming.

 

[Mr Davis 97]

So I think I might have an answer. First, we note that any element of <a^k> is also an element of <a>, by the definition of <a>. So <a^k> is a subset of <a>.

Second, we need to show that <a> is a subset of <a^k>, that is, any element of <a> is also an element of <a^k>.
First we take note that n, the order of the group, and k, are relatively prime. That is, there exist integers b and c s.t. bk + cn = 1, which implies that for any integer m, (mb)k + (mc)n = m.
Take an arbitrary element of <a>, $a^m$. Then $a^m = a^{(mb)k + (mc)n} = a^{(mb)k} a^{(mc)n} = (a^{mb})^{k} (a^{n})^{mc} = (a^{mb})^{k} e = (a^{mb})^{k}$. Hence, any element of <a> is also an element of <a^k>. Thus, <a> is a subset of <a^k>.

Thus, <a> = <a^k>, when k is relatively prime to the order of the group.

Is this right?

 

[STF92]

In fact, when (k,n)=1 where n is the order of <a>, a^k is a generator of <a>, by the same reasoning you did. So, <a>= <a^k> automatically. It is equivalent to the fact that the group of the integers module n has [k] for a generator for any k such that (n,k)=1.

 

[fresh_42]

Looks correct, although it would be better to write $(a^{mb})^k=(a^k)^{mb}$ because you want to have powers of $a^k$.