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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 $|G|$.

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  • $\begingroup$ It seems that $G$ being abelian is also used to define the surjective morphism in the prime case, no? Otherwise, I don't see why the given map should necessarily be a homomorphism at all. $\endgroup$ – Alex Wertheim Jul 25 '16 at 6:28
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    $\begingroup$ The prime case is also well known in the non-abelian case via Cauchy's theorem. $\endgroup$ – MooS Jul 25 '16 at 6:29
  • $\begingroup$ @AlexWertheim Ouch, you're right. Thanks! But... wait, I'm gonna edit my question. $\endgroup$ – Sora. Jul 25 '16 at 6:30
  • $\begingroup$ @MooS One minute faster, that's just the edit I wanted to make. $\endgroup$ – Sora. Jul 25 '16 at 6:31
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Yes, the argument of Case 2 is valid for any Dedekind group.

Note that you have to replace the argument in Case 1 by Cauchy's theorem in the non-abelian case.

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In general, converse of Lagrange theorem holds for any finite nilpotent group (Proof) and any finite Dedekind group is nilpotent.

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  • $\begingroup$ It even holds for any finite supersolvable group, see here. $\endgroup$ – Dietrich Burde Mar 25 '18 at 11:06
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Some facts: Dedeking groups are of the form $Q\times B\times D$ where $Q$ is the quaternion groups, $B$ is a boolean group (elementary $2$-group), and $D$ is an odd order torsion group. The latter can be very complicated, but in the finite case it is just a product of cyclic groups. These facts are not trivial.

So, the answer to your question is yes, every Dedekind group has the reverse Lagrange property, since it is a product of groups each having the reverse Lagrange property.

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