Showing $\{g \in G: g^n=1\}$ is a cyclic subgroup of cyclic $G$ of order $n$ for $|G|=nk, k\in\Bbb N$.

Question

Show that $H=\{g \in G: g^n=1\}$ is a cyclic subgroup of cyclic group $G$ of order $n$ where $|G|=d$ and $d=nk$ for some $k \in \Bbb N$.

I could have used Cauchy's theorem but then I realised that this would be an overkill.

So

1. I proved that every subgroup of a cylic group is cyclic.
2. Then I proved that $H\leq G$ so $H$ is cyclic by (1). $1 \in H$ so $H \neq \phi$ and $g^{k}(\neq 1)\in H$. So, $|H|\geq 2$.

Now I am stuck at the point on how to prove that $|H|=n$ in an elementary way? Any help?

Note: it's not a duplicate of this as there the question was to prove that it subgroup which I did by myself. Moreover, that is a question on commutative group and asked to prove that it is a subgroup and here the question is on order of the subgroup.

Answer 1

Since $H$ is cyclic, $H = \langle h\rangle$ for some $h \in G$ with $h^n = 1$. Thus the order of $H$ must divide $n$. Conversely, if $g$ is a generator of $G$ (i.e. $G = \left<g\right>$), then $g^{ik}$ is in $H$ for every $i = 0,1,...,n-1$ since $(g^{ik})^n = g^{id} = 1$. Since each $g^{ik}$ is distinct, there are at least $n$ elements of $H$. The only number greater than or equal to $n$ that divides $n$ is $n$, so $|H| = n$.