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

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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

Proof

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