# 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 $1 \le l \le k_j$ and $1 \le i,j \le r$.

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

## 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 $1 \le l \le k_j$ and $1 \le i,j \le r$. $G$ 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 $1 \le l \le k_j$ Step (3) [SHOW MORE]