I'm currently reading a book on algebra and the following argument is used to prove that the converse of Lagrange's theorem (if $d$ divides $|G|$ there exists a subgroup of order $d$) holds for any finite Abelian group:
Proof by mathematical induction. For $|G|\leq 3$ the statement is trivial.
Case 1: $d=p$ is a prime number and $$G':=\prod_{g\in G}\langle g\rangle\to G,\quad(g_1,\cdots,g_n)\mapsto g_1\cdots g_n$$ is a surjective homomorphism, so $|G|$ divides $|G'|=\prod_{g\in G}\operatorname{ord}g$, so $\operatorname{ord}g=p\cdot k$ for some $g\in G,k\in\mathbb Z$, in which case $\langle g^k\rangle$ is a subgroup of order $p$.Okay, so of course this map is surjective, but not a homomorphism, if $G$ is not Abelian. But by Cauchy's theorem, we still know that there exists a subgroup of order $p$ for any prime divisor $p$ of $|G|$.
Case 2: $d$ is arbitrary. Let $p$ be a prime factor of $d$, so by case 1 we have a subgroup $H'$ of order $p$, so $G/H'$ is a group and $|G/H'|<|G|$, so by the induction hypothesis, there exists a subgroup $\bar H\subseteq G/H'$ of order $d':=d/p$, so the preimage $\pi^{-1}(\bar H)$ of the canonical projection $\pi:G\to G/H'$ is a subgroup of $G$ of order $d$.
I understand the fact that $G$ is Abelian was used when we stated that $G/H'$ is a group. But is there any reason this proof would not hold for any finite Dedekind group?
On a second thought: Of course, if we know a group we're examining is a Dedekind group, we probably know all of its subgroups so we wouldn't need this statement, but still, does it hold? For example, could we state the following exercise?
Let $G$ be finite group such that all subgroups are normal. Prove that there exists a subgroup of order $d$ for any divisor of $|G|$.
