Recently, I was looking at a property possessed by some finite groups - a finite group, $G$, has the property (call it property $P$) if, for any $g_1,g_2,\dots,g_n\in G$ s.t. $G=\langle g_1,g_2,\dots,g_n\rangle$, we have that $|G|$ divides $\prod_{i=1}^n \operatorname{ord}(g_i)$.
The reason I found this interesting initially was because it was a property that all finite abelian groups possess but one that only some finite nonabelian groups possess.
$S_6$ and $\operatorname{Dih}_8$ (octagon) are examples of nonabelian groups that lack the property (more generally all symmetric groups and all dihedral groups can easily be shown to lack property $P$).
It also turns out that the quaternion group, $Q_8$, possesses property $P$. This made me suspect that perhaps property $P$ related to Dedekind groups.
One can, in fact, easily prove (more generally than the abelian case) that all finite Dedekind groups possess property $P$.
However, I do not know if the condition that a finite group be Dedekind is equivalent to it possessing property $P$ (I think, however, that this will not be true).
Can anyone provide some examples of groups that satisfy property $P$ but are not Dedekind? (specifically, examples that may shed some light on common properties of these $P$ property groups and how they may be further classified)
Or is there some reason to believe that there are no such examples?
Links to existing relevant literature, if there is any, would also be very appreciated.
(Proof for Abelian case: If $G$ is any abelian group and $G=\langle g_1,g_2,\dots,g_n\rangle$, then the function $h:\mathbb Z_{\operatorname{ord}(g_1)}\times \mathbb Z_{\operatorname{ord}(g_2)}\times \dots \times \mathbb Z_{\operatorname{ord}(g_n)}\rightarrow G$ defined as $h(k_1,k_2,\dots,k_n)=g_1^{k_1}g_2^{k_2}\dots g_n^{k_n}$ is a surjective homomorphism, so by the first isomorphism theorem, we can conclude that $|G|$ divides $|\mathbb Z_{\operatorname{ord}(g_1)}\times \mathbb Z_{\operatorname{ord}(g_2)}\times \dots \times \mathbb Z_{\operatorname{ord}(g_n)}|=\prod_{i=1}^n \operatorname{ord}(g_i)$, so that $G$ possesses property $P$.
Proof for general Dedekind case: The trivial group is Dedekind and it clearly possesses property $P$. Now, let $G$ be a group and suppose all Dedekind groups of order $<|G|$ are known to possess property $P$. Suppose also that $G=\langle g_1,g_2,\dots,g_n\rangle$. Suppose WLOG that $g_1\neq e$. Since $G$ is Dedekind, $\langle g_1\rangle \trianglelefteq G$, so the group $G/\langle g_1\rangle$ is well-defined. Since $g_1\neq e$, $G/\langle g_1\rangle$ has order $<|G|$ - since it is a homomorphic image of $G$, it too is Dedekind. This means $G/\langle g_1\rangle$ possesses property $P$. Since $G=\langle g_1,g_2,\dots,g_n\rangle$, $G/\langle g_1\rangle=\langle g_2\langle g_1\rangle,g_3\langle g_1\rangle,\dots,g_n\langle g_1\rangle\rangle$ - by the induction hypothesis, then $|G/\langle g_1\rangle|$ divides $\prod_{i=2}^n \operatorname{ord}(g_i\langle g_1\rangle)$. This in turn divides $\prod_{i=2}^n \operatorname{ord}(g_i)$ so we conclude that $|G|$ divides $\prod_{i=1}^n \operatorname{ord}(g_i)$ as desired.)
Edit: A way of constructing new groups that have property $P$ is to the direct product of any two groups, $A$ and $B$, such that each of $A$ and $B$ have property $P$, and $|A|$ and $|B|$ are coprime (easy to prove).
Let $H$ be a finite group and let $n$ such that $2^n\geq |H|$. Then $G:=Q_8^n\times H$ is such a group.
Proof: Any generating set of $G$ must project to a generating set of (the quotient) $Q_8^n$. Now, $\Phi(Q_8^n)=C_2^n$ and $Q_8^n/\Phi(Q_8^n)\cong C_2^{2n}$ so, by Burnside's Basis Theorem, we need $2n$ elements to generate $Q_8^n$ and, moreover, every element outside $\Phi(Q_8^n)$ has order $4$, so the product of the element orders is at least $4^{2n}=16^n$. This is clearly true also for the original generating set (of $G$). Since $2^n\geq |H|$, $16^n\geq |G| $.
In particular, any group can appear as a normal subgroup. To me, this suggests that there is no meaningful classification of these groups.
EDIT: Sorry, I just noticed that you require the order to "divide" not just be less or equal. So this argument only works for $2$-groups.
I'm curious now if there are any non-nilpotent examples, or if the odd Sylows are always abelian?