Past exam of the building infinite groups course of the University of Oxford / 2011 / 1

Use the commutator convention $[x,y]=x^{-1}{y}^{-1}xy$, and let $[A,B]$ denote the subgroup generated by all such commutators with $x\in A,y\in B$. The lower central series isA central series has $[G_j,G]\le G_{j-1}$ for each $j\ge 1$. We prove the requested inclusion by induction. At $i=0$, ${\gamma }_1(G)=G_n$. If $i<n$ and ${\gamma }_{i+1}(G)\le G_{n-i}$, thenThereforeIn particular ${\gamma }_{n+1}(G)=1$. The proof uses the central-series condition directly; no equality of the intermediate terms is claimed.

Suppose inductively that $G=H{\gamma }_i(G)$. In the quotient $G/{\gamma }_{i+1}(G)$, the image of ${\gamma }_i(G)$ is central. Hence this quotient is generated by the image of $H$ together with a central subgroup, so its commutator subgroup lies in the image of $H$. Pulling this back givesThe original assumption now gives $G=H{\gamma }_2(G)\le H{\gamma }_{i+1}(G)\le G$. Starting with $i=2$, induction proves $G=H{\gamma }_i(G)$ for every $i\ge 2$. These products are subgroups because each ${\gamma }_i(G)$ is normal; no normality of $H$ is needed.

If $G$ is finitely generated, every quotient, including its abelianization $G/{\gamma }_2(G)$, is finitely generated. Conversely, lift a finite generating set of that quotient to $g_1,\dots ,g_t\in G$, and set $H=⟨g_1,\dots ,g_t⟩$. Then $G=H{\gamma }_2(G)$. For a nilpotent group, ${\gamma }_{c+1}(G)=1$ for some $c$, and the established identity gives $G=H$. Thus

All commutators of $H$ with $K$ lie in the central subgroup $Y$ of $HK$. In the standard commutator identities, conjugation by elements of $H$ or $K$ therefore acts trivially on these commutators. Since $Y$ is abelian, their order can also be reversed. ConsequentlyThus the map is a homomorphism in each variable separately; this does not assert that it is a homomorphism from the direct product jointly.

It factors through $H/C_H(K)$ in the first variable and $K/C_K(H)$ in the second. If these finite quotient groups have representatives $h_1,\dots ,h_a$ and $k_1,\dots ,k_b$, then the finite collection $[h_i,k_j]$ generates $Y$. Let $e$ be an exponent of $H/C_H(K)$. For every $h,k$,Hence $Y$ is a finitely generated abelian group of bounded exponent, and $Y$ is finite. For instance its order is at most $e^{ab}$.

Now suppose $G$ is nilpotent and $G/{\gamma }_2(G)$ is finite. Induct on the nilpotency class $c$. For $c=1$, $G$ is its own abelianization and is finite. If $c\ge 2$, the group $\bar{G}=G/{\gamma }_c(G)$ has smaller class andsince ${\gamma }_c(G)\le {\gamma }_2(G)$. By induction $\bar{G}$ is finite. Apply the preceding result with $H=G={\gamma }_1(G)$ and $K={\gamma }_{c-1}(G)$. Their commutator subgroup is $Y={\gamma }_c(G)\le Z(G)$. Both relevant centralizer quotients are finite, because their centralizers contain ${\gamma }_c(G)$ and $G/{\gamma }_c(G)$ is finite. Thus ${\gamma }_c(G)$ is finite. The extension of this finite normal subgroup by the finite quotient $\bar{G}$ is finite. The converse follows because a quotient of a finite group is finite. Therefore