# Classification of abelianness-forcing numbers

$n>1$. Prove that $n$ is an abelianness-forcing number iff. $n=p_1^{a_1}p_2^{a_2}\dots p_r^{a_r}$, where $p_1,p_2,\dots,p_r$ are distinct primes, is

• -cubefree
• -$p_i\nmid p_j^{a_j}-1,\forall i,j$

(Abstract Algebra: Dummit & Foote, Semidirect products)

The 'only if' part requires me to construct non-abelian groups of order $n$ if $n$ fails to be cubefree or $p_i\mid p_j^{a_j}-1$, for some $i,j$.

If $n$ is divisible by $p^3$ for some prime $p$, we can have either $\Bbb Z_{p^2}\rtimes\Bbb Z_p$ or $(\Bbb Z_p\times\Bbb Z_p)\rtimes\Bbb Z_p$ as a direct product factor for a group of order $n$. Surely it can't be abelian.

If $n$ has two distinct prime factors $p$ and $q$. If for instance, $n=pq\dots$ and $p\mid q-1$, we can have $\Bbb Z_q\rtimes\Bbb Z_p$ as a factor for a group of order $n$. Again it can't be abelian. If $n=pq^2\dots$ and $p\mid q^2-1$, which in this case doesn't imply $p\mid q-1$, we have $(\Bbb Z_q\times\Bbb Z_q)\rtimes\Bbb Z_p$ as a direct product factor. The homomorphism associated with this semidirect product can be nontrivial: $$\varphi:\Bbb Z_p\rightarrow\operatorname{Aut}(\Bbb Z_q\times\Bbb Z_q)\cong GL_2(\Bbb F_q)$$ $GL_2(\Bbb F_q)$ has order $(q^2-1)(q^2-q)$, divisible by $p$, so it has a subgroup of order $p$. Again we're done.

I have no idea how to deal with the 'if' part, however. The authors do not much hint. They only refer us back to the exercise showing that 'if $G$ is a finite group, with every proper subgroup being abelian, then $G$ is solvable'$-(*)$. I decide to use induction. Let $G$ be a group with an order that is cubefree and satisfies the divisibility condition for all prime factors. Assume the result is true for all groups with order $<|G|$. If $H$ is any proper subgroup of $G$, then its order is cubefree and satisfies the divisibility condition. By induction, $H$ is abelian. By $(*)$, $G$ is solvable. Derived series is either: $1\lhd G$, or $1\lhd [G,G]\lhd G$. I don't know how to show that the latter case is impossible. I think I'm also far from finishing the proof... Can somebody please help?

• Hint: Consider a $p$-Sylow subgroup of $G/[G,G]$ for some prime $p$, and consider its preimage in $G$. Show that if this is a proper subgroup of $G$ then the $p$-Sylow subgroup of $G$ is normal. – Tobias Kildetoft Aug 2 '17 at 13:37
• @TobiasKildetoft I don't get it. How does this relate to the proof? And apparently, I have no idea how to begin. – user441558 Aug 2 '17 at 14:54
• You need to show that all Sylow subgroups are normal. My hint allows you to show this for most of them. – Tobias Kildetoft Aug 2 '17 at 15:16
• @DerekHolt While the marked duplicate is about the same result, there does not actually seem to be any proof of the result there. – Tobias Kildetoft Aug 2 '17 at 17:18
• As a further hint, once you have found a normal Syow $p$-subgroup $P$, we have $|P| \le p^2$, and the assumptions tell you that $|G/P|$ is coprime to ${\rm Aut}(P)|$, so $P \le Z(P)$. Now you can show that all Sylow subgroups are normal. – Derek Holt Aug 3 '17 at 8:11