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Is there a classification of finite groups $G$ with the property that any prime dividing $|G|$ must also divide $|G^{ab}|$ (the order of its abelianization)?

Being nilpotent is sufficient, though not necessary. On the other hand, being supersolvable is not sufficient, since the abelianization of any dihedral group $D_{2k}$ is a 2-group, and is also not necessary.

Are there other interesting families of finite groups which have this property?

(In particular I'm interested in groups generated by 2 elements)

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    $\begingroup$ The group $G$ need not be solvable. One example is $G = A_{5} \times \mathbb{Z}/30\mathbb{Z}$. $\endgroup$ – Jeremy Rouse Oct 25 '16 at 17:41
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    $\begingroup$ Or similarly, for ay finite group $H$ of order $n,$ the group $G = H \times \left( \mathbb{Z}/n\mathbb{Z} \right)$ would have order $n^{2}$ and Abelianization of order divisible by $n$, so there is no hope of a classification without further restrictions. $\endgroup$ – Geoff Robinson Oct 25 '16 at 17:49
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    $\begingroup$ As a note - Jeremy Rouse's group can be generated by 2 elements. $A_5$ can be generated by an element $a$ of order 3 and an element $b$ of order 5, and $\mathbb{Z}/30\mathbb{Z}$ can be generated by 5 and 6. Then $(a, 6), (b, 5)$ generate $(0, 18), (0, 25), (a^2, 0), (b, 0)$, and so generate the whole group. More generally, any finite group $H$ generated by two elements of coprime order leads to $G = H \times (\mathbb{Z}/n\mathbb{Z})$ satisfying your condition in a similar manner. $\endgroup$ – user44191 Oct 25 '16 at 20:09
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    $\begingroup$ @user44191 it's a bit heuristic, but because it's hard, if not practically impossible, to give nice families of finite groups such that admit all finite groups as quotients... except trivialities such as "the family of all finite groups" or variants. By "nice" family I would mean a list of "nice groups". [In contrast to the fact that every finite group embeds into some alternating group, for instance.] $\endgroup$ – YCor Oct 26 '16 at 5:14
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    $\begingroup$ When $H$ is a finite group and $n$ is just the product of all primes which divide $|H|$ but not $|H^{ab}|$, then $G= H\times C_n$ has the property in question. If $H$ is $2$-generated, then so is $G$. $\endgroup$ – Frieder Ladisch Oct 26 '16 at 10:05

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