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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.

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  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 1 at 0:20
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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.

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