Classification of nilpotency-forcing numbers

Statement

The following are equivalent for a natural number $n$:

1. Every group of order $n$ is a nilpotent group (and in particular, a finite nilpotent group)
2. $n$ has prime factorization of the form $n=p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{r}^{k_{r}}$ where $p_{i}$ does not divide $p_{j}^{l}-1$ for any $1\leq l\leq k_{j}$ and $1\leq i,j\leq r$.

The result is sometimes also called Schmidt's theorem. It is a corollary of the Schmidt-Iwasawa theorem.

Related facts

• Classification of cyclicity-forcing numbers
• Classification of abelianness-forcing numbers

Facts used

1. Lagrange's theorem
2. Schmidt-Iwasawa theorem: This states that any group all of whose proper subgroups are nilpotent must be solvable; in particular, it must have a nilpotent maximal normal subgroup such that the quotient group is of prime order.
3. Equivalence of definitions of finite nilpotent group
4. Characteristic of normal implies normal

Proof

(1) implies (2)

It suffices to prove the contrapositive. In other words, we show that if the condition in (2) is violated, then we can construct a non-nilpotent group of that order.

Given: A natural number $n$, distinct primes $p,q$ and a natural number $l$ such that $pq^{l}$ divides $n$, and $p$ divides $q^{l}-1$.

To prove: There is a non-nilpotent group of order $n$

Proof: Note that the product $pq^{l}$ divides $n$.

We can construct a non-nilpotent group $H$ of order $pq^{l}$ as follows: consider the additive group of the field of size $q^{l}$. The multiplicative group of this field is a cyclic group of size $q^{l}-1$. Since $p$ divides $q^{l}-1$, it has a subgroup of order $p$. Construct a semidirect product of the additive group of order $q^{l}$ with this subgroup of order $p$.

We can now construct a non-nilpotent group $G$ of order $n$ as the external direct product:

$G=H\times \mathbb {Z} /(n/pq^{l})\mathbb {Z}$

(2) implies (1)

We prove the claim by induction on the number. Note that any divisor of a number that satisfies the condition also satisfies the condition.

Base case for induction ($n=1$): Obvious

Inductive step: The inductive hypothesis is that the result holds for all smaller numbers.

Given: A natural number $n$ has prime factorization of the form $n=p_{1}^{k_{1}}p_{2}^{k_{2}}\dots p_{r}^{k_{r}}$ where $p_{i}$ does not divide $p_{j}^{l}-1$ for any $1\leq l\leq k_{j}$ and $1\leq i,j\leq r$. $G$ is a group of order $n$.

To prove: $G$ is nilpotent

Proof:

Step no.	Assertion/construction	Facts used	Give data used	Previous steps used	Explanation
1	Every proper subgroup of $G$ is nilpotent	Fact (1)	inductive hypothesis arithmetic condition on $n$		Any divisor of $n$ satisfies the conditions necessary for the inductive hypothesis to apply. Applying the inductive hypothesis, we obtain that every subgroup is nilpotent.
2	Either $G$ is nilpotent or it contains a nilpotent maximal normal subgroup $N$ such that the quotient group $G/N$ is cyclic of prime order $p$ for some $p$ dividing $n$	Fact (2)		Step (1)	Step-fact combination direct
3	$G$ has a nilpotent normal $p$-complement $K$.	Facts (3), (4)		Step (2)	$G$ has a nilpotent normal subgroup of index $p$. This in turn has normal Sylow subgroups for all primes dividing its order, which are characteristic in it and therefore also normal in $G$. For all primes other than $p$, order considerations make these Sylow subgroups of $G$ as well.
4	Suppose $P$ is a $p$-Sylow subgroup of $G$. Then, every element of $P$ centralizes all elements of $K$. Thus, $G$ is an internal direct product of $P$ and $K$, and is nilpotent.		$p_{i}$ does not divide $p_{j}^{l}-1$ for $1\leq l\leq k_{j}$	Step (3)	[SHOW MORE] Suppose $x\in P$ induces a nontrivial automorphism of $K$ by conjugation. Since $K$ is nilpotent, this restricts to an automorphism on its Sylow subgroups, and must be nontrivial on at least one of them. Suppose it is nontrivial on a Sylow subgroup $P_{j}$ of order $p_{j}^{k_{j}}$, $p_{j}\neq p$. The automorphism has order a power of $p$, so pick a power of it that has order exactly $p$. Call this automorphism $\sigma$. Thus, $\sigma$ is a nontrivial automorphism of $P_{j}$ of order $p$. Denote by $M$ the fixed-point subgroup of $P_{j}$ under $\sigma$. $M$ has order $p_{j}^{m}$ for some $m<k_{j}$. The orbit structure under $\sigma$ for the remaining $p_{j}^{k_{j}}-p_{j}^{m}=p_{j}^{m}(p_{j}^{k_{j}-m}-1)$ elements comprises orbits of size $p$. Thus, $p$ divides $p_{j}^{k_{j}-m}-1$. However, this contradicts the assumption that $p$ cannot divide $p_{j}^{l}-1$ for $1\leq l\leq k_{j}$. Therefore, it is not possible for elements of $P$ to induce nontrivial automorphisms of $K$. Thus, every element of $P$ centralizes every element of $K$, and the conclusion follows.