Assume that $G$ is a cyclic group of order $n$, that $G=\langle a \rangle$, that $k\mid n$ and that $H=\langle a^k \rangle.$ Find $[G:H]$.

Question

Assume that $G$ is a cyclic group of order $n$, that $G=\langle a \rangle$, that $k\mid n$ and that $H=\langle a^k \rangle.$ Find $[G:H]$.

Ok, so I have to find $[G:H]$ which is equal to $\frac {|G|}{|H|}$ which equals $\frac {n}{\langle a^k \rangle}$ but I'm not sure what to do after that.

Answer 1

No, $$\begin{align}[G:H]&=\frac{|G|}{|H|} \\ &=\frac{n}{n/k} \\ &=k.\end{align}$$

Answer 2

I'll complete this result for the case when $k$ doesn't necessarily divide $n$. Then, $ |H|=\dfrac n{\gcd(n,k)}$, and therefore $$[G:H]=|G/H|=\gcd(n,k).$$