Lower central series $\gamma_{n+1}(G)$ are smallest normal subgroups of $G$ such that $\gamma_n(G)/\gamma_{n+1}(G) \leq Z(G/\gamma_{n+1}(G))$

For a group $$G$$ and subgroups $$H,K$$ define $$(H,K)$$ to be the group generated by the commutators $$(h,k)$$ where $$h \in H, k \in K$$.

Define the lower central series of $$G$$ as:

• $$\gamma_1(G) = G$$
• $$\gamma_{n}(G) = (G, \gamma_{n-1}(G)), \; \forall n \geq 2$$

I am asked to show that the lower central series $$\gamma_{n+1}(G)$$ is the smallest normal subgroup of $$G$$ such that $$\gamma_n(G)/\gamma_{n+1}(G) \leq Z(G/\gamma_{n+1}(G))$$

I am assuming this means that for every $$n$$, $$\gamma_{n+1}(G)$$ is minimal with respect to inclusion, in the set $$\{K \triangleleft G, \gamma_n(G) \mid \gamma_n(G)/K \leq Z(G/K)\}$$

Given this, I suspect that I should try to show that for any other such normal subgroup $$K$$ of $$G \cap \gamma_n(G) = \gamma_n(g)$$, then $$\gamma_{n+1}(G) \leq K$$

To this end: let $$g \in G, z \in \gamma_n(G), k \in K$$. Then $$(g,z) \in \gamma_{n+1}(G)$$ and:

$$g^{-1}kg = z^{-1}kz = k$$ as $$K$$ normal in $$\gamma_n(G), G$$

$$\rightarrow g^{-1}z^{-1}kzg = k$$

$$\rightarrow (g,z)zgkzg = k$$

$$\rightarrow (g,z) = g^{-1}z^{-1}k^{-1}gzk = g^{-1}z^{-1}k^{-1}zgg^{-1}z^{-1}gzk$$

$$\rightarrow (g,z) = k^{-1}(g,z)k$$

So we see that $$(g,z)$$ and $$k$$ commute, and hence $$\gamma_{n+1}(G)$$ and $$K$$ commute. However, now I'm stuck and don't really know how to keep going to show that $$\gamma_{n+1}(G) \leq K$$.

Any help would be appreciated, thank you.

• I’m not sure how you get that $g^{-1}kg=z^{-1}kz$, especially not for a random element of $k$. You are asserting this holds for any $k\in K$, and that is certainly incorrect. – Arturo Magidin Apr 1 at 0:20

Suppose that $$N$$ is a (normal) subgroup of $$G$$ contained in $$\gamma_n(G)$$ such that $$\gamma_n(G)/N\leq Z(G/N)$$. We want to show that every element of $$\gamma_{n+1}(G)$$ must lie in $$N$$. For that, it suffices to show that the generating set of $$\gamma_{n+1}(G)$$ lies in $$N$$. To that end let $$x\in\gamma_n(G)$$ and $$g\in G$$. We want to show that the commutator $$(g,x)\in N$$.
Now, since $$\gamma_n(G)/N\leq Z(G/N)$$, then $$xN$$ is central in $$G/N$$; therefore, it commutes with $$gN$$. But $$xN$$ commutes with $$gN$$ if and only if the commutator $$(gN,xN)$$ in $$G/N$$ is trivial. Since $$(gN,xN) = (g,x)N$$, this holds if and only if $$(g,x)\in N$$. Thus, $$(g,x)\in N$$, as desired.
Thus, we have shown that every generator of $$\gamma_{n+1}(G)$$ lies in $$N$$, so $$\gamma_{n+1}(G)\leq N$$, as desired.