Cyclics groups and Lagrange Theorem 
Let $G=\langle y\rangle$ be a cyclic group of order $n$. Prove that $\langle y^k\rangle \subseteq\langle y^m\rangle $ if and only if $\gcd(n,m)$ divides $\gcd(n,k)$.

So I think I was able to prove one direction. I argued that:
Since $<y^k>\subseteq <y^m>$ and $<y^m>$ is a cyclic group, we know that $<y^k>$ is a cyclic subgroup of $<y^m>$. This implies via lagrange theorem that the order of $<y^m>$ must be divisible by the order of $<y^k>$. The order of $<y^m>$ equals $\frac{n}{gcd(n,m)}$ and the order of $<y^k>$ equals $\frac{n}{gcd(n,k)}$. We know that $\frac{|<y^m>|}{|<y^k>|}=\frac{\frac{n}{gcd(n,m)}}{\frac{n}{gcd(n,k)}}=\frac{gcd(n,k)}{gcd(n,m)}$. Hence, gcd(n,m) divides gcd(n,k).
I was having difficulties in trying to prove the other side. Any suggestions or help?
Thank You.
 A: To get the other direction, since $(n,m)\mid (n,k)$, (I use the notation $(n,k)$ for $\gcd(n,k)$) let
$$(n,k)=(n,m)d$$
for some integer $d$ and since $(n,k)\mid k$, let
$$k=(n,k)c$$
for some integer $c$. Lastly, recall that we can write
$$(n,m)=\alpha n+\beta m$$
for some integers $\alpha,\beta$. Then, for any $(y^{k})^a=y^{ak}\in\langle y^k\rangle$, we get
$$y^{ak}=y^{a(n,k)c}=y^{acd(n,m)}=y^{acd(\alpha n+\beta m)}=(y^n)^{\alpha acd}(y^m)^{\beta acd}$$
but $G$ has order $n$ so $y^n=1$. Hence,
$$(y^k)^a=(y^m)^{\beta acd}\in\langle y^m\rangle$$
That is,
$$\langle y^k\rangle\subseteq\langle y^m\rangle$$
A: The main point is indeed that the order of $y^{k}$, and thus of the subgroup it spans, is 
$$
\frac{n}{\gcd(n, k)}.\tag{formula}
$$
Then your statement follows from the fact that $G$ has exactly one subgroup for each divisor of its order, so $$
\langle y^ {k} \rangle \le \langle y^ {m} \rangle \text{ iff } \frac{n}{\gcd(n, k)} \text{ divides } \frac{n}{\gcd(n, m)} \text{ iff } \gcd(n, m) \text{ divides }\gcd(n, k).
$$
To prove (formula), note that if $(y^{k})^{t} = 1$, with $t > 0$, then $n$ divides $kt$, and thus $n/\gcd(n, k)$ divides $t$.
