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

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