I am trying to understand the proof of the following theorem.
Theorem: If $G$ is a finite abelian group and $d$ is a divisor of $|G|$, then $G$ contains a subgroup of order $d$.
Proof: Let $d$ be any divisor of $|G|$, and let $p$ be a prime divisor of $d$. From Cauchy's theorem, it follows that there is $S\leq G$ of order $p$. $S$ must be normal since $G$ is abelian, and hence $G/S$ is an abelian group of order $n/p$. By induction on $|G|$, $G/S$ has a group $H^*$ of order $d/p$. From the Correspondence Theorem, $H^*=H/S$ for some subgroup $H$ of $G$ containing $S$, and $|H|=|H^*||S|=d$.
I totally understand the big picture of this proof. However, I don't understand why $G/S$ has a group $H^*$ of order $d/p$? Based on my understanding, the inductive assumption should be that the theorem holds for any group $G$ of order not greater $n$. Therefore, by induction on $|G|$, there exists a subgroup $H^*$ of order $d$. Why there is a subgroup of order $d/p$?
Forgive me if this question is too dumb, but I really think I just miss something very obvious.