Proving o(An) = o(a) for Finite Abelian Groups | G, N, a | Group Theory Homework

[Justabeginner]

Homework Statement

Let G be a finite group with N , normal subgroup of G, and a, an element in G.
Prove that if (a) intersect N = (e), then o(An) = o(a).

Homework Equations

The Attempt at a Solution

(aN)^(o(a)) = a^(o(a)) * N = eN = N, but is the least power such that (aN)^m = N. Assume m must divide o(a).

(aN)^((o(a)) = (aN)^ (mq +r) where 0 <= r < m,
However, ((aN)^m)-q * a(N)^(o(a)) = (a(N)^r)= N = (a(N)^r).
r < m so r= 0 and mq= o(a).
I am not sure how to continue however, am I even going in the right direction?

Thanks!

 

[gopher_p]

Homework Statement

Let G be a finite group with N , normal subgroup of G, and a, an element in G.
Prove that if (a) intersect N = (e), then o(An) = o(a).

Homework Equations

The Attempt at a Solution

(aN)^(o(a)) = a^(o(a)) * N = eN = N, but is the least power such that (aN)^m = N. Assume m must divide o(a).

(aN)^((o(a)) = (aN)^ (mq +r) where 0 <= r < m,
However, ((aN)^m)-q * a(N)^(o(a)) = (a(N)^r)= N = (a(N)^r).
r < m so r= 0 and mq= o(a).
I am not sure how to continue however, am I even going in the right direction?

Thanks!

Presumably $(a)$ is meant to indicate the cyclic subgroup of $G$ generated by the element $a$, $aN$ is the left coset $aN=\{an:n\in N\}$ considered as an element of the quotient group $G/N$, $o(a)$ is order of $a$ as an element of $G$ and $o(aN)$ the order of the element $aN$ in the quotient group $G/N$.

You've correctly identified that $aN^{o(a)}=a^{o(a)}N=eN=N$ and concluded (I think) that $o(aN)|o(a)$, or at the very least that $o(aN)\leq o(a)$ (which is all that's really needed).

What remains to be shown is that $o(a)\leq o(aN)$. This is where you probably want to use the necessary fact that $(a)\cap N=(e)$.

Hint:

Spoiler

For all $k$, $a^k\in aN^k$

 

[Justabeginner]

That made sense and I think I got it. Thank you very much!