# Number of nilpotent groups equals product of number of groups of order each maximal prime power divisor

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Statement

Suppose $n$ is a natural number with prime factorization:

$n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r}$

where $p_1,p_2,\dots,p_r$ are distinct primes and $k_1,k_2,\dots,k_r$ are natural numbers.

Then the number of nilpotent groups of order $n$ is the product of the number of groups of order $p_i^{k_i}$ for all $1 \le i \le r$:

(Number of groups of order $p_1^{k_1}$) $\times$ (Number of groups of order $p_2^{k_2}$) $\times$ $\dots$ $\times$ (Number of groups of order $p_r^{k_r}$)

## Facts used

1. Equivalence of definitions of finite nilpotent group