My attempt
Note- Addition is under $mod$ $d$
Let $G = (a^0,a^1,a^2.....a^{n-1})$
$n=kd$
Let $H_d = (a^{0k},a^{1k},a^{2k}......a^{{(d-1)}k})$
For $a^{z_1k},a^{z_2k} \in H_d$
$a^{z_1k}a^{z_2k} = a^{{(z_1+z_2)}k} \in H_d$
$H_d$ is closed
For $a^{zk} \in H_d$
$a^{zk} (a^{zk})^{-1}=a^0$
$(a^{zk})^{-1} = (a^{-1})^{zk}$
Since $a^{-1}$ is $a^{n-1}$ we can rewrite it as
$(a^{zk})^{-1} = (a^{n-1})^{zk}=a^{z(n-1)k} \in H_d$
Since $(a^{zk})^{-1} \in H_d$ and $H_d$ is closed, $H_d$ is a subgroup.
I can easily prove the 3rd part by using Lagrange's theorem.
I have no idea how to prove that only $1$ subgroup of order $d$ exists. Maybe by proving that the elements of the $2$ groups have to be the same?
I would really appreciate if someone could verify my proof so far and give some hints to complete the proof.
To show a subgroup $H$ of order $d$ is unique, observe that $b^d$ must be the identity for each $b\in H$. Write $b=a^k$, with $0\le k<n$. Then $b^d=a^{kd}$, and for this to equal the identity, $kd$ must be a multiple of $n$, that is $kd/n$ is an integer. This $k$ is a multiple of $n/d$, so that $b$ is an element of the set $\{a^0,a^{n/d},a^{2n/d},\ldots,a^{n-n/d}\}$. But this set has exactly $d$ elements so $H$ must equal $\{a^0,a^{n/d},a^{2n/d},\ldots,a^{n-n/d}\}$.