Recursive expression for $\Lambda(n)$: number of nilpotent groups of order $n < \infty$ My question is related to this one here about nilpotent groups, but it is different since I want to prove an explicit expression for the number of nilpotent groups.

Suppose  $n = p_1^{k_1}p_2^{k_2}\ldots p_m^{k_m}$, where $p_i$ is a prime number for every $1\le i\le m$. If $\Lambda(n)$ denotes the number of nilpotent groups of order $n$, (how) can we prove that$$\Lambda(n) = \Lambda(p_1^{k_1})\Lambda(p_1^{k_1})\ldots \Lambda(p_m^{k_m})$$
If not, can we make a similar statement?

Here's what I've thought so far:
Let $G$ be a nilpotent group of order $n = p_1^{k_1}p_2^{k_2}\ldots p_m^{k_m}$. Let $S_i$ denote the unique $p_i$-Sylow subgroup of $G$. Since $G$ is nilpotent, we can write $G$ as the following direct product:
$$G \cong S_1\times S_2\times \ldots\times S_m$$
How do I proceed from here? I'm unable to see the recursive structure if any. Also, I have the following theorem with me if it helps:

Let $|G| = n$ and $n = p_1^{k_1}p_2^{k_2}\ldots p_m^{k_m}$. Then the following are equivalent:

*

*$G$ is nilpotent.

*If $H < G$ (notation for subgroup), then $H < N_G(H)$.

*Every Sylow subgroup is normal.

*$G \cong S_1\times S_2\times \ldots\times S_m$, where $S_i$ denotes the unique $p_i$-Sylow subgroup of $G$.


Would appreciate any help! Thanks a lot.
 A: If $G$ is nilpotent then as you did we can write $G\cong S_1\times S_2 \times...\times S_m$ where $|S_i|={p_i}^{k_i}$, so the problem of counting all the nilpotent groups of order $n$ boils down to counting all groups of the form $S_1\times S_2 \times...\times S_m$  where $|S_i|={p_i}^{k_i}$ (up to isomorphism). We know that every $p$-group is nilpotent so the number of nilpotent groups of order ${p_i}^{k_i}$ is just the number of groups of that order, also, every choice of $S_1,S_2,...,S_m$ defines a uniqe group in the sence that if $L_1,...,L_m$ are also groups of order $|L_i|={p_i}^{k_i}$ and $S_1\times S_2 \times...\times S_m \cong L_1\times L_2 \times...\times L_m$ then $S_i\cong L_i$ since they must have the same sylow subgroups.
All in all, there are $\Lambda ({p_i}^{k_i})$ groups of order ${p_i}^{k_i}$ for each $i$.  Every choice of $m$ such groups defines a unique nilpotent group of order $n$, and every nilpotent group of order $n$ is isomorphic to a group formed this way so you can conclude
$\Lambda(n) = \Lambda ({p_1}^{k_1})...\Lambda ({p_m}^{k_m})$.
