How to understand Solvable and Nilpotent Groups

I hope to give a natural explanation for the following phenomena by viewing the terms of the derived and lower central series as subfunctors of the identity on groups.

1. Why is each term $D^n(G)$$D^n(G)$ or ${\gamma }_n(G)$$\gamma_n(G)$ normal in the original group $G$$G$?

2. Why do both series commute with finite direct products? Does every subfunctor of the identity on groups share this property?

3. Why do the quotients $G/D^{\ell }(G)$$G/D^\ell(G)$ and $G/{\gamma }_{c+1}(G)$$G/\gamma_{c+1}(G)$ give left adjoints to the inclusions of the corresponding bounded classes? How do their values on free groups determine the two Lawvere theories?

4. How do these theories explain why a fixed bound on derived length or nilpotency class is preserved under subgroups, quotients, and small limits?

5. Why are solvable groups closed under extensions? What part of the argument applies to the iterates of any subfunctor of the identity?

The derived subgroup as a kernelTwo recursions on subfunctorsFinite productsEach term as the kernel of a reflectionThe two Lawvere theories from their free groupsSubgroups, quotients and limitsClosure under extensions

 

The derived subgroup as a kernel

Let $I:\mathsf{Ab}\hookrightarrow \mathsf{Grp}$$I:\mathsf{Ab}\hookrightarrow\mathsf{Grp}$ denote the inclusion of abelian groups into groups, and let

$$\eta :{\mathrm{Id}}_{\mathsf{Grp}}\Rightarrow I\circ \mathrm{Ab}$$

be the unit of the abelianization adjunction $\mathrm{Ab}⊣I$$\mathrm{Ab}\dashv I$. Taking the kernel of this natural transformation defines the derived subgroup functor

$$D:=\ker (\eta ).$$

Kernels in the functor category are computed pointwise, so for each group $G$$G$ we have

$$D(G)=\ker ({\eta }_G:G⟶G^{\mathrm{ab}}).$$

To identify this kernel, observe that every homomorphism from $G$$G$ to an abelian group kills every commutator $xy{x}^{-1}{y}^{-1}$$xyx^{-1}y^{-1}$. The subgroup $[G,G]$$[G,G]$ generated by these commutators is normal, since conjugation sends commutators to commutators. The quotient $G/[G,G]$$G/[G,G]$ is abelian, and every homomorphism from $G$$G$ to an abelian group factors uniquely through it. Thus $G/[G,G]$$G/[G,G]$ has the universal property of the abelianization, and

$$D(G)=[G,G].$$

Taking the kernel in the functor category gives us a natural inclusion

$$i:D\hookrightarrow {\mathrm{Id}}_{\mathsf{Grp}}.$$

For every homomorphism $f:G\to H$$f:G\to H$, naturality says that

$$f\circ i_G=i_H\circ D(f).$$

In elementwise terms, $f(D(G))\subseteq D(H)$$f(D(G))\subseteq D(H)$, and $D(f):D(G)\to D(H)$$D(f):D(G)\to D(H)$ is the restriction of $f$$f$.

Two recursions on subfunctors

For subgroups $A,B\le G$$A,B\leq G$, write $[A,B]$$[A,B]$ for the subgroup generated by all $[a,b]$$[a,b]$ with $a\in A$$a\in A$ and $b\in B$$b\in B$. For normal $A$$A$ and $B$$B$, each generating commutator lies in both, so $[A,B]\le A\cap B$$[A,B]\leq A\cap B$. We can take this bracket pointwise on subfunctors of the identity.

Lemma (bracket of subfunctors). If $K,L\hookrightarrow {\mathrm{Id}}_{\mathsf{Grp}}$$K,L\hookrightarrow\mathrm{Id}_{\mathsf{Grp}}$ are subfunctors, then

$$[K,L](G):=[K(G),L(G)]$$

defines a subfunctor of the identity. The construction is monotone in both arguments.

Proof. For any homomorphism $f:G\to H$$f:G\to H$,

$$f([K(G),L(G)])=[f(K(G)),f(L(G))]\le [K(H),L(H)].$$

The equality follows by applying $f$$f$ to the generating commutators. The inclusion follows from naturality of $K$$K$ and $L$$L$. Monotonicity follows because enlarging either subgroup enlarges the set of generating commutators. $◻$$\square$

Now put $D^0={\gamma }_1={\mathrm{Id}}_{\mathsf{Grp}}$$D^0=\gamma_1=\mathrm{Id}_{\mathsf{Grp}}$ and define the two series by

$$D^{k+1}=[D^k,D^k],{\gamma }_{k+1}=[{\gamma }_k,{\mathrm{Id}}_{\mathsf{Grp}}].$$

The first recursion applies for $k\ge 0$$k\geq 0$, the second for $k\ge 1$$k\geq 1$. Starting from the identity, the lemma shows inductively that every term is a subfunctor. In the derived series the superscript records genuine functor composition, since $[D^k(G),D^k(G)]=D(D^k(G))$$[D^k(G),D^k(G)]=D(D^k(G))$. In the lower central series the second argument remains the original group $G$$G$. We have $D={\gamma }_2$$D=\gamma_2$, and the next terms are

$$D^2(G)=[[G,G],[G,G]],{\gamma }_3(G)=[[G,G],G].$$

A group is solvable if $D^{\ell }(G)=1$$D^\ell(G)=1$ for some $\ell \ge 0$$\ell\geq 0$, and nilpotent if ${\gamma }_{c+1}(G)=1$$\gamma_{c+1}(G)=1$ for some $c\ge 0$$c\geq 0$. The least such $\ell$$\ell$ and $c$$c$ are its derived length and nilpotency class, respectively. Thus $D^2(G)=1$$D^2(G)=1$ says that the commutator subgroup is abelian, while ${\gamma }_3(G)=1$$\gamma_3(G)=1$ says that it lies in the center of $G$$G$.

Why is every term normal in the original group? Consider any subfunctor $K\hookrightarrow {\mathrm{Id}}_{\mathsf{Grp}}$$K\hookrightarrow\mathrm{Id}_{\mathsf{Grp}}$. Fix a group $G$$G$ and apply naturality to an endomorphism $f:G\to G$$f:G\to G$. We obtain

$$f(K(G))\subseteq K(G).$$

Thus $K(G)$$K(G)$ is fully invariant in $G$$G$. In particular, conjugation by any $g\in G$$g\in G$ restricts to an automorphism of $K(G)$$K(G)$, with inverse given by conjugation by $g^{-1}$$g^{-1}$. Hence $K(G)⊴G$$K(G)\trianglelefteq G$.

Then we apply this observation to $K=D^k$$K=D^k$ and $K={\gamma }_k$$K=\gamma_k$.

Corollary. Every nilpotent group is solvable.

Proof. We have $D^k\le {\gamma }_{k+1}$$D^k\leq\gamma_{k+1}$ for all $k\ge 0$$k\geq 0$. The two sides agree at $k=0$$k=0$. If the inclusion holds at $k$$k$, monotonicity of the bracket gives

$$D^{k+1}=[D^k,D^k]\le [{\gamma }_{k+1},{\mathrm{Id}}_{\mathsf{Grp}}]={\gamma }_{k+2}.$$

In particular, ${\gamma }_{c+1}(G)=1$$\gamma_{c+1}(G)=1$ implies $D^c(G)=1$$D^c(G)=1$. $◻$$\square$

This bound is far from sharp. The inclusion $[{\gamma }_i,{\gamma }_j]\le {\gamma }_{i+j}$$[\gamma_i,\gamma_j]\leq\gamma_{i+j}$, a consequence of the three subgroups lemma, improves the comparison to $D^k\le {\gamma }_{2^k}$$D^k\leq\gamma_{2^k}$. A group of class $c$$c$ therefore has derived length at most $\lceil {\log }_2(c+1)\rceil$$\lceil\log_2(c+1)\rceil$.

Finite products

Again let $K\hookrightarrow {\mathrm{Id}}_{\mathsf{Grp}}$$K\hookrightarrow\mathrm{Id}_{\mathsf{Grp}}$ be any subfunctor. To see what the two projections tell us, take an element $(g,h)\in K(G\times H)$$(g,h)\in K(G\times H)$. Naturality of the inclusion means that the projection $p_G:G\times H\to G$$p_G:G\times H\to G$ restricts to a homomorphism

$$K(p_G):K(G\times H)⟶K(G).$$

Thus $g=p_G(g,h)$$g=p_G(g,h)$ belongs to $K(G)$$K(G)$. Applying the same argument to the second projection $p_H:G\times H\to H$$p_H:G\times H\to H$ gives $h\in K(H)$$h\in K(H)$. Both coordinates therefore lie in their respective subgroups, so

$$K(G\times H)\subseteq K(G)\times K(H).$$

For the reverse inclusion, take $x\in K(G)$$x\in K(G)$ and $y\in K(H)$$y\in K(H)$. Naturality for the coordinate inclusions puts both $(x,1)$$(x,1)$ and $(1,y)$$(1,y)$ in $K(G\times H)$$K(G\times H)$. Since this is a subgroup, it contains their product $(x,y)$$(x,y)$. Hence

$$K(G\times H)=K(G)\times K(H).$$

Since $K(1)=1$$K(1)=1$ as well, every subfunctor of the identity on $\mathsf{Grp}$$\mathsf{Grp}$ preserves finite products.

In particular,

$$\begin{array}{rl}{D}^k(X\times Y)& =D^k(X)\times D^k(Y),\\ {\gamma }_k(X\times Y)& ={\gamma }_k(X)\times {\gamma }_k(Y)(k\ge 1).\end{array}$$

A direct product of two groups is solvable if and only if both factors are solvable, and then its derived length is the maximum of their derived lengths. The same argument gives the corresponding statement for nilpotency and the maximum of the two classes. In either case, a given term of the product is trivial exactly when the corresponding terms of both factors are trivial.

Each term as the kernel of a reflection

Write ${\mathsf{Sol}}_{\le \ell }$$\mathsf{Sol}_{\leq\ell}$ for the groups of derived length at most $\ell$$\ell$, and ${\mathsf{Nil}}_{\le c}$$\mathsf{Nil}_{\leq c}$ for the groups of nilpotency class at most $c$$c$, each with all group homomorphisms between its objects. At bound zero both categories consist of trivial groups, and at bound one both are $\mathsf{Ab}$$\mathsf{Ab}$. We shall use $n$$n$ for the arity of an operation.

Let $K$$K$ be either $D^{\ell }$$D^\ell$ or ${\gamma }_{c+1}$$\gamma_{c+1}$, and write $\mathcal{V}$$\mathcal V$ for the corresponding full subcategory of groups $H$$H$ satisfying $K(H)=1$$K(H)=1$. The quotient map

$${\eta }_G^K:G\twoheadrightarrow G/K(G)$$

is universal among homomorphisms from $G$$G$ to an object of $\mathcal{V}$$\mathcal V$. There are two things to check.

First, the quotient belongs to $\mathcal{V}$$\mathcal V$. For a surjection $q:G\twoheadrightarrow Q$$q:G\twoheadrightarrow Q$, the equality $q([A,B])=[q(A),q(B)]$$q([A,B])=[q(A),q(B)]$ applies at every step of either recursion. Starting with $q(G)=Q$$q(G)=Q$, induction gives

$$q(D^k(G))=D^k(Q),q({\gamma }_k(G))={\gamma }_k(Q)$$

for their respective indices. Apply this to $q={\eta }_G^K$$q=\eta^K_G$ to get $K(G/K(G))=1$$K(G/K(G))=1$.

Second, if $f:G\to H$$f:G\to H$ and $K(H)=1$$K(H)=1$, then $f(K(G))\le K(H)=1$$f(K(G))\leq K(H)=1$ by naturality. Hence $f$$f$ factors uniquely through ${\eta }_G^K$$\eta^K_G$.

Thus $R_K(G)=G/K(G)$$R_K(G)=G/K(G)$ defines a functor left adjoint to the inclusion $I_{\mathcal{V}}:\mathcal{V}\hookrightarrow \mathsf{Grp}$$I_{\mathcal V}:\mathcal V\hookrightarrow\mathsf{Grp}$, and $K=\ker ({\eta }^K)$$K=\ker(\eta^K)$ as a subfunctor of the identity. Taking $K=D={\gamma }_2$$K=D=\gamma_2$ recovers abelianization. Composing the reflector with the free group functor gives the free object in each bounded class. On $n$$n$ generators these are

$$F_n/D^{\ell }(F_n)\text{and}{F}_n/{\gamma }_{c+1}(F_n),$$

where $F_n$$F_n$ is the free group on $x_1,\dots ,x_n$$x_1,\ldots,x_n$.

The two Lawvere theories from their free groups

Fix a derived length bound $\ell$$\ell$ or a nilpotency class bound $c$$c$. In either case, write $\mathcal{V}$$\mathcal V$ for the bounded class and $A_n=F_n/K(F_n)$$A_n=F_n/K(F_n)$ for its free group on $n$$n$ generators.

Theory	$\mathcal{V}$$\mathcal V$	$A_n$$A_n$
${\mathbb{T}}_{\ell }^{\mathrm{sol}}$$\mathbb T^{\mathrm{sol}}_\ell$	${\mathsf{Sol}}_{\le \ell }$$\mathsf{Sol}_{\leq\ell}$	$F_n/D^{\ell }(F_n)$$F_n/D^\ell(F_n)$
${\mathbb{T}}_c^{\mathrm{nil}}$$\mathbb T^{\mathrm{nil}}_c$	${\mathsf{Nil}}_{\le c}$$\mathsf{Nil}_{\leq c}$	$F_n/{\gamma }_{c+1}(F_n)$$F_n/\gamma_{c+1}(F_n)$

Define the theory $\mathbb{T}$$\mathbb T$ in either row to have objects $n\ge 0$$n\geq 0$ and arrows

$$\mathbb{T}(n,m)={\mathrm{Hom}}_{\mathsf{Grp}}(A_m,A_n)\cong A_n^m.$$

The bijection records the images of the $m$$m$ free generators. Thus an arrow $n\to m$$n\to m$ is an $m$$m$-tuple of words in $n$$n$ generators, taken modulo $K(F_n)$$K(F_n)$, and composition is substitution. In particular, $n$$n$ is the $n$$n$-fold product of $1$$1$, with projections given by the generator classes ${\overline{x}}_1,\dots ,{\overline{x}}_n$$\bar x_1,\ldots,\bar x_n$. The object $0$$0$ is terminal.

Proposition. Evaluation at $1$$1$ identifies the category of set-valued models of $\mathbb{T}$$\mathbb T$ with $\mathcal{V}$$\mathcal V$ up to equivalence.

Proof. Given $G\in \mathcal{V}$$G\in\mathcal V$, define

$$M_G(n)={\mathrm{Hom}}_{\mathsf{Grp}}(A_n,G)\cong G^n.$$

An arrow represented by $A_m\to A_n$$A_m\to A_n$ acts by precomposition. Under the displayed bijection the projections act as coordinate projections, so $M_G$$M_G$ preserves finite products and $M_G(1)\cong G$$M_G(1)\cong G$.

Conversely, let $M:\mathbb{T}\to \mathsf{Set}$$M:\mathbb T\to\mathsf{Set}$ preserve finite products and put $G=M(1)$$G=M(1)$, identifying $M(n)$$M(n)$ with $G^n$$G^n$. The quotient maps $F_n\to A_n$$F_n\to A_n$ respect substitution and hence give a morphism from the theory of groups to $\mathbb{T}$$\mathbb T$. Restricting $M$$M$ along it makes $G$$G$ a group. To see which bound this group satisfies, choose any tuple $\mathbf{g}=(g_1,\dots ,g_n)\in G^n$$\mathbf g=(g_1,\ldots,g_n)\in G^n$ and evaluate the operations at that tuple:

$$e_{\mathbf{g}}:A_n⟶G,a⟼M(a)(\mathbf{g}).$$

This is a group homomorphism. For example, the operation $ab$$ab$ is the composite of $(a,b):n\to 2$$(a,b):n\to 2$ with multiplication, so $e_{\mathbf{g}}(ab)=e_{\mathbf{g}}(a)e_{\mathbf{g}}(b)$$e_{\mathbf g}(ab)=e_{\mathbf g}(a)e_{\mathbf g}(b)$. Since ${\overline{x}}_i$$\bar x_i$ is the $i$$i$th projection, $e_{\mathbf{g}}({\overline{x}}_i)=g_i$$e_{\mathbf g}(\bar x_i)=g_i$. Its image is therefore exactly $H=⟨g_1,\dots ,g_n⟩$$H=\langle g_1,\ldots,g_n\rangle$. Applying the surjection formula from the preceding section to $A_n\twoheadrightarrow H$$A_n\twoheadrightarrow H$ gives $K(H)=e_{\mathbf{g}}(K(A_n))=1$$K(H)=e_{\mathbf g}(K(A_n))=1$.

Every finitely generated subgroup of $G$$G$ thus satisfies the same bound. To pass to $G$$G$, write it as the directed union of these subgroups. Both recursions give

$$K(G)=\bigcup _{\begin{array}{c}H\le G\\ H\mathrm{finitely}\mathrm{generated}\end{array}}K(H).$$

Indeed, at each bracket step an element is a finite product of commutators and their inverses. By induction all its inputs occur at the preceding steps in finitely many such subgroups, which lie in a single finitely generated subgroup. The reverse inclusion is naturality. Hence $K(G)=1$$K(G)=1$.

Finally, evaluation at $\mathbf{g}$$\mathbf g$ is now the unique homomorphism $A_n\to G$$A_n\to G$ sending ${\overline{x}}_i$$\bar x_i$ to $g_i$$g_i$. Thus $M$$M$ and $M_G$$M_G$ agree on every operation under $M(n)\cong G^n$$M(n)\cong G^n$. A natural transformation between models is determined by its component at $1$$1$, which preserves the group operations. Conversely, every group homomorphism induces a natural transformation by postcomposition. This gives

$$\mathrm{Mod}({\mathbb{T}}_{\ell }^{\mathrm{sol}},\mathsf{Set})\simeq {\mathsf{Sol}}_{\le \ell },\mathrm{Mod}({\mathbb{T}}_c^{\mathrm{nil}},\mathsf{Set})\simeq {\mathsf{Nil}}_{\le c}.$$

$◻$$\square$

Subgroups, quotients and limits

Subgroups and quotients. If $K(G)=1$$K(G)=1$ and $H\le G$$H\leq G$, then $K(H)\le K(G)$$K(H)\leq K(G)$ by naturality. If $q:G\twoheadrightarrow Q$$q:G\twoheadrightarrow Q$, then $K(Q)=q(K(G))$$K(Q)=q(K(G))$. Hence subgroups and quotients preserve the bound in either theory.

Small limits at a fixed bound. Fix either row of the table, and let $(G_j{)}_{j\in J}$$(G_j)_{j\in J}$ be a small diagram of its models. Limits of models are computed pointwise. Indeed, if $M_j$$M_j$ is the model associated with $G_j$$G_j$, then

$$\left(\underset{j}{lim}{M}_j\right)(n)\cong \underset{j}{lim}{M}_j(1{)}^n\cong {(\underset{j}{lim}{M}_j(1))}^n.$$

Thus the pointwise limit preserves finite products and its underlying group is the ordinary limit in $\mathsf{Grp}$$\mathsf{Grp}$. One can also see the bound directly. If $L=\underset{j}{lim}{G}_j$$L=\lim_j G_j$ with projections $p_j$$p_j$, then $p_j(K(L))\le K(G_j)=1$$p_j(K(L))\leq K(G_j)=1$ for every $j$$j$. The projections are jointly monic, so $K(L)=1$$K(L)=1$.

Finite diagrams and arbitrary products. A nonempty finite diagram of solvable groups admits a common bound on their derived lengths. Its limit is therefore solvable. Taking the maximum of the nilpotency classes gives the same conclusion for a finite diagram of nilpotent groups. The empty limit is the trivial group. For any small diagram a common finite bound still suffices. For a direct product it is also necessary, because every factor is a quotient of the product. Explicitly,

$$\begin{array}{rl}\prod _j{G}_j\text{ is solvable}& ⟺\mathrm{\exists }\ell \ge 0\mathrm{\forall }j,D^{\ell }(G_j)=1,\\ \prod _j{G}_j\text{ is nilpotent}& ⟺\mathrm{\exists }c\ge 0\mathrm{\forall }j,{\gamma }_{c+1}(G_j)=1.\end{array}$$

For the solvable case, the free objects already supply a family of unbounded derived length. The group $G_{\ell }:=F_2/D^{\ell }(F_2)$$G_\ell:=F_2/D^\ell(F_2)$ has derived length exactly $\ell$$\ell$. To see this, note that $F_2$$F_2$ is not solvable, since it surjects onto every two-generator group, including $A_5$$A_5$. Thus each $D^k(F_2)$$D^k(F_2)$ is nontrivial. It is also free by the Nielsen–Schreier theorem, and a nontrivial free group has nontrivial abelianization. Hence $D^{k+1}(F_2)⊊D^k(F_2)$$D^{k+1}(F_2)\subsetneq D^k(F_2)$ for every $k$$k$, and for $\ell \ge 1$$\ell\geq 1$

$$D^{\ell -1}(G_{\ell })=D^{\ell -1}(F_2)/D^{\ell }(F_2)\ne 1.$$

For the nilpotent case, take the dihedral group $P_t$$P_t$ of a regular $2^t$$2^t$-gon, with $t\ge 2$$t\geq 2$. A rotation $r$$r$ of order $2^t$$2^t$ and a reflection $s$$s$ generate $P_t$$P_t$, with $sr{s}^{-1}=r^{-1}$$srs^{-1}=r^{-1}$. The formula $[r^a,s]=r^{2a}$$[r^a,s]=r^{2a}$ gives

$${\gamma }_k(P_t)=⟨r^{2^{k-1}}⟩(k\ge 2).$$

Indeed, the quotient by $⟨r^2⟩$$\langle r^2\rangle$ is abelian and $[r,s]=r^2$$[r,s]=r^2$, which gives the first step. Rotations commute, and every reflection inverts $r$$r$, so each subsequent step doubles the exponent. Thus $P_t$$P_t$ has class $t$$t$. Each $P_t$$P_t$ also has abelian derived subgroup, so $\prod _{t\ge 2}{P}_t$$\prod_{t\geq 2}P_t$ has derived length two and is not nilpotent. For the lower central series of dihedral groups in general, see Example 4.16 of Keith Conrad, Subgroup Series I.

These examples give countable products outside the respective classes, so neither unbounded class can be defined by identities in the group signature. For a general limit the arrows also matter. Arrange either family in an inverse sequence with every transition map trivial. Every compatible tuple is then trivial, so its inverse limit has both derived length and nilpotency class zero.

Closure under extensions

The two recursions part ways on extensions. The derived series iterates a single subfunctor, and that is exactly what the following argument uses.

Proposition. Let $K\hookrightarrow {\mathrm{Id}}_{\mathsf{Grp}}$$K\hookrightarrow\mathrm{Id}_{\mathsf{Grp}}$ be a subfunctor, and write $K^j$$K^j$ for its $j$$j$-fold iterate. The class of groups $G$$G$ for which $K^j(G)=1$$K^j(G)=1$ for some $j$$j$ is closed under extensions.

Proof. Consider an extension $1\to N\to G\stackrel{q}{\to }Q\to 1$$1\to N\to G\xrightarrow{q}Q\to 1$ with $K^a(N)=1$$K^a(N)=1$ and $K^b(Q)=1$$K^b(Q)=1$. By naturality, $q(K^b(G))\le K^b(Q)=1$$q(K^b(G))\leq K^b(Q)=1$, so $K^b(G)\le N$$K^b(G)\leq N$. Applying $K^a$$K^a$ to this inclusion gives

$$K^{a+b}(G)=K^a(K^b(G))\le K^a(N)=1.$$

$◻$$\square$

Taking $K=D$$K=D$ recovers the extension bound for derived length. An extension of solvable groups is solvable, with derived length at most $a+b$$a+b$.

The proof rests on the composition law $K^{a+b}=K^a\circ K^b$$K^{a+b}=K^a\circ K^b$. The derived series has it, because $D^k$$D^k$ is the $k$$k$-fold iterate of $D$$D$. The lower central series does not. Its recursion keeps the second argument at the whole group, so the terms ${\gamma }_k$$\gamma_k$ are not the successive iterates of a single subfunctor. Starting from ${\gamma }_{b+1}(G)\le N$$\gamma_{b+1}(G)\leq N$, the same argument only gives

$${\gamma }_{a+1}({\gamma }_{b+1}(G))\le {\gamma }_{a+1}(N)=1.$$

This controls commutators taken inside ${\gamma }_{b+1}(G)$$\gamma_{b+1}(G)$, whereas the later terms of the lower central series of $G$$G$ take commutators with all of $G$$G$.

The conclusion fails as well. In the extension

$$1⟶A_3⟶S_3⟶C_2⟶1$$

both $A_3$$A_3$ and $C_2$$C_2$ are abelian. A transposition inverts each $3$$3$-cycle, so $[A_3,S_3]=A_3$$[A_3,S_3]=A_3$ and ${\gamma }_k(S_3)=A_3$$\gamma_k(S_3)=A_3$ for every $k\ge 2$$k\geq 2$. Thus $S_3$$S_3$ is not nilpotent. It has derived length two, which also shows that the bound $a+b$$a+b$ can be attained.

What survives for the lower central series is the central case. If $N$$N$ is central in $G$$G$ and ${\gamma }_{c+1}(G/N)=1$$\gamma_{c+1}(G/N)=1$, then ${\gamma }_{c+1}(G)\le N$$\gamma_{c+1}(G)\leq N$ and

$${\gamma }_{c+2}(G)=[{\gamma }_{c+1}(G),G]\le [N,G]=1.$$

So the choice made at the second step, commutators within the subgroup or commutators with the whole group, is also the choice between closure under all extensions and closure under central extensions.