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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.