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If $G$ is an arbitrary group we denote its abelianization as $G^{ab} := \frac{G}{[G,G]}$, where $[G,G]$ is the commutator. As an abelian group it is characterized by the following universal property:

Let $\pi : G \to G^{ab}$ be the quotient morphism. If $\varphi : G \to H$ is an morphism between $G$ and an abelian group $H$, then there is a unique morphism $\hat{\varphi} : G^{ab} \to H$ such that $\varphi = \hat{\varphi}\circ\pi$.

In a sense the abelianization is the abelian group that is closest to $G$. I wonder if there is a solution for the analogous universal property, but with nilpotent groups instead of abelian. (Or, perhaps only over nilpotent groups of class $n$ for fixed $n\in\mathbb{N}$, the abelianization being the case $n=1$)

J. W. Tanner
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Eric Vaz
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    You have to fix the class $n$, otherwise free groups provide a counter-example to the existence of a universal nilpotent quotient. – Moishe Kohan Dec 28 '23 at 02:56
  • These are answers to the question. Comments should only be used to clarify, not answer the question. See How do comments work for more information. – Martin Brandenburg Dec 28 '23 at 03:41
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    @MarianoSuárez-Álvarez Well, one could argue that you answered the question and then turned it into a question by appending "no?". But despite what the rules may say about comments, they are useful in practice for providing quick hints and suggestions, which is what you are doing. – Derek Holt Dec 28 '23 at 08:46
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    It is also worth pointing out that, although not all groups have a largest nilpotent quotient, many do, including all finite groups. There is an algorithm due to Charles Sims that computes and proves correctness of the largest nilpotent group of a group defined by a finite presentation if it exists. So in particular it can be used to verify that the presentation defines a nilpotent group. – Derek Holt Dec 28 '23 at 08:51

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For any variety of groups $\mathfrak{V}$ (a class of groups that is closed under subgroups, arbitrary products, and homomorphic images), given a group $G$ there is a verbal subgroup $\mathfrak{V}(G)$ with the property that:

  1. $\mathfrak{V}(G)$ is normal in $G$ (in fact, fully invariant);
  2. $G/\mathfrak{V}(G)\in \mathfrak{V}$;
  3. Any morphism from $G$ to a group in $\mathfrak{V}$ factors through $G/\mathfrak{V}(G)$.

The subgroup $\mathfrak{V}(G)$ is generated by all values of words that are laws of the variety $\mathcal{V}$. See the discussion here.

The class of all nilpotent groups of class at most $c$ is determined by the law $$[[\cdots[[x_1,x_2],x_3]\cdots],x_{c+1}]$$ (that is, a group $G$is nilpotent of class at most $c$ if and only if every evaluation of that word at elements of $G$ yields the identity element). The corresponding verbal subgroup is also known as the $(c+1)$st term of the lower central series of $G$ and plays the same role relative to nilpotent groups of class at most $c$ as the commutator subgroup (second term of the lower central series) relative to abelian groups (nilpotent groups of class at most $1$).

However, there is no similar subgroup for the class of all nilpotent groups. You can take the intersection of all terms of the lower central series, which will have properties 1 and 3 above, but need not have property 2, so it cannot be called a "nilpotenciation" of $G$. Similar things happen with "solvable of length at most $n$" (use the $n$th term of the derived series) vs. "solvabilization".

(Some groups may have a subgroup satisfying 1, 2, and 3 relative to "all nilpotent groups"; if the terms of the lower central series stabilize, then that subgroup is the one you want. E.g., all finite groups necessarily have this property; for instance, the "nilpotenciation" of $S_3$ is its abelianization).

Arturo Magidin
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