Let $N \unlhd G$. Then $G/N$ is nilpotent of class $c\in\Bbb{N}$ iff $c$ is the smallest natural number such that $\gamma_c(G) \subset N$

Question

I think I have established the following proposition:

Theorem: Let $G$ be a group and $N \unlhd G$. Then, $G/N$ is nilpotent of class $c \in \mathbb{N}$ if, and only if, $c$ is the smallest natural number such that $\gamma_c(G) \subset N$

Note: $\gamma_0(G) = G$, $\gamma_1(G) = [G, G] = G’$ and so on

Proof

$\implies$) Suppose $G/N$ is nilpotent of class $c$. Then $c$ is the smallest natural number such that $\gamma_c(G/N) = 1$. But then $1 = \gamma_c(G)N/N$, which implies $\gamma_c(G)N \subset N$, meaning $\gamma_c(G) \subset N$. The minimality of $c$ comes from a similar argument with $\gamma_{c-1}(G/N)$

$\impliedby$) Now suppose $c$ is the smallest natural number such that $\gamma_c(G) \subset N$. This means that $\gamma_c(G)N \subset N$, which, in turn, implies $\gamma_c(G)N/N = 1$. Therefore, $\gamma_c(G/N) = 1$. Again, we can conclude that $c$ is the class of $G/N$ by using a similar argument with $c-1$. $\square$

Is the above correct?

While I don’t really see any mistakes, this seems like a quite straightforward generalization of the famous “$G/N$ is abelian iff $G’ \subset N$”, and yet I have never seen this result before. This gave me some uncertainty as to its validity…

Thanks in advance!

Answer 1

As I noted in the comments, your numbering of the terms of the lower central series is not the usual one. We usually define $\gamma_1(G)=G$, and $\gamma_{n+1}(G) = [G_n,G]$ for $n\geq 1$, so that a group is nilpotent of class at most $c$ if and only if $G_{c+1}=\{e\}$ (and of class exactly $c$ if $G_{c+1}=\{e\}$ and $G_c\neq\{e\}$).

Other than that, your result and argument is fine. It is a special case of a more general result.

If you look carefully at your argument, you will see that you are not using much about the lower central series (in particular, you are not really using its precise definition). The facts that you are using are:

1. Every group $K$ has a subgroup $\gamma_c(K)$.
2. If $\phi\colon G\to M$ is a group homomorphism, then $\phi(\gamma_c(G))\leq \gamma_c(M)$, and if $\phi$ is surjective, then we have equality.

So more generally, let us consider a construction $\mathbf{V}$ with properties 1 and 2 above: for every group $K$ we have a subgroup $\mathbf{V}(K)$, the construction is functorial, so that $\phi(\mathbf{V}(G))\leq\mathbf{V}(M)$ for any morphism $\phi\colon G\to M$; and if $\phi$ is surjective then we have equality. Let $\mathfrak{V}$ be the collection of all groups $G$ for which $\mathbf{V}(G)=\{e\}$.

Proposition. Let $G$ be a group, and let $N\triangleleft G$. Then $G/N\in\mathfrak{V}$ if and only if $\mathbf{V}(G)\leq N$.

Proof. Let $\pi\colon G\to G/N$ be the canonical morphism.

If $G/N\in\mathfrak{V}$, then $\pi(\mathbf{V}(G))\leq \mathbf{V}(G/N)=\{e\}$, so $\mathbf{V}(G)\leq N$.

Conversely, if $\mathbf{V}(G)\leq N$, then $\mathbf{V}(G/N) = \pi(\mathbf{V}(G)) = \mathbf{V}(G)N/N = N/N = \{e\}$, so $G/N\in\mathfrak{V}$, as claimed. $\Box$

What kind of constructions $\mathbf{V}$ satisfy properties 1 and 2? A very natural one is derived from varieties of groups.

Roughly speaking a variety of groups can be characterized two different ways:

1. As a class of groups that is closed under taking homomorphic images, subgroups, and arbitrary direct products. That is, if $G$ is in the class, then so is every subgroup and homomorphic image of $G$, and every arbitrary direct product of groups in the class is again in the class.

2. As the class of all groups that satisfy a set of laws; equivalently, given a subset $S$ of the free group $F_{\omega}$ of countably infinite rank, groups $G$ with the property that given any homomorphism $\phi\colon F_{\omega}\to G$, we have $\phi(S)=\{e\}$.

For example, class of all abelian groups is a variety, and it is determined by the law $[x,y]=x^{-1}y^{-1}xy$ (think of all groups for which all evaluations of $x^{-1}y^{-1}xy$ yield $e$). The class of all groups of exponent $7$ is a variety, determined by the law $x^7$. And so on.

The class of nilpotent groups of class at most $c$ is determined by the law $[x_1,x_2,\ldots,x_c,x_{c+1}]$ (with commutators left-normed, so that $[a,b,c]=[[a,b],c]$).

Given any variety of groups $\mathfrak{V}$, for every group $G$ there is a least normal subgroup $\mathbf{V}(G)\triangleleft G$ such that $G/\mathbf{V}(G)\in \mathfrak{V}$ (you can prove this by noting that there is at least one such normal subgroup, and that the intersection of any family of such normal subgroups is again such a normal subgroup). This is called the verbal subgroup of $G$ associated to $\mathfrak{V}$. An intrinsic way of defining it is to identify a set $S$ of laws that determine $\mathfrak{V}$, and letting $\mathbf{V}(G)$ be the subgroup of $G$ generated by all values that the elements of $S$ take when evaluated at elements of $G$; that is, the subgroup $\langle \phi(S)\mid \phi\colon F_{\omega}\to G\text{ is a homomorphism}\rangle$.

Proposition. Let $\mathfrak{V}$ be a variety determined by the set of laws $S$. Then $\mathbf{V}$ is a construction that satisfies properties 1 and 3 above.

Proof. By definition, $\mathbf{V}(K)$ is a subgroup of $K$ for every group $K$. If $\phi\colon G\to M$ is a group homomorphism, then for every morphism $\psi\colon F_{\omega}\to G$ we obtain a morphism $\phi\circ\psi\colon F_{\omega}\to M$, so that $\phi(\psi(S))\subseteq \mathbf{V}(M)$. Thus, $\phi(\langle \psi(S)\mid \psi\colon F_{\omega}\to G\rangle)\leq \mathbf{V}(M)$.

Finally, assume that $\phi$ is surjective. The subgroup $\mathbf{V}(M)$ is generated by elements obtained by taking $s\in S$ and "evaluating" it at elements of $M$. That is, if $s=s(x_1,\ldots,x_k)$ is an element of $F_{\omega}$, we select elements $m_1,\ldots,m_k\in M$ (not necessarily distinct) and consider the value of $s(m_1,\ldots,m_k)$. Take one such element. Since $\phi$ is surjective, there exist $g_1,\ldots,g_k\in G$ such that $\phi(g_i)=m_i$. It is now straightforward to verify that $$s(m_1,\ldots,m_k) = s(\phi(g_1),\ldots,\phi(g_k)) = \phi(s(g_1,\ldots,g_k)).$$ Since $s(g_1,\ldots,g_k)\in\mathbf{V}(G)$, it follows that $s(m_1,\ldots,m_k)\in \phi(\mathbf{V}(G))$, proving that $\mathbf{V}(M)\leq\phi(\mathbf{V}(G))$ when $\phi$ is surjective, and hence giving equality. $\Box$

The result you have proven follows by considering first the variety of nilpotent groups of class at most $c$, which shows that $G/N$ is nilpotent of class at most $c$ if and only if $\gamma_{c+1}(G)\leq N$ (using the usual nomenclature). Then we note that $G/N$ is not of class at most $c-1$ if and only if $\gamma_c(G)\nleq N$. Combining the two we get that $G/N$ is nilpotent of class exactly $c$ if and only if (i) $\gamma_{c+1}(G)\leq N$ and (ii) $\gamma_c(G)\nleq N$. Thus, if and only if $\gamma_{c+1}(G)$ is the first term of the lower central series of $G$ that is contained in $N$.

A similar result holds for "solvable of solvability length exactly $c$ (you require that the appropriate term of the derived series be contained in $N$; "solvable of length $1$" corresponds to the law $[x,y]$; of length $2$ corresponds to the law $[[x_1,x_2],[x_3,x_4]]$, etc.). And any variety of groups.