Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 3 / 3

For a finitely generated group $G=⟨x_1,\dots ,x_r⟩$, every quotient group is generated by the images of the $x_i$. Thus finite generation passes to quotients.

For a group extension $1\to N\to G\to Q\to 1$, suppose $N$ and $Q$ are finitely generated. Choose finite group generators of $N$ and lifts to $G$ of finite group generators of $Q$. For any $g\in G$, a word in the lifts has the same image as $g$ in $Q$, so their difference belongs to $N$. Thus the two finite sets together generate $G$, proving preservation under extensions.

For a subgroup $H\le G$ of finite index, choose representatives $T$ of the right cosets $H\backslash G$, with $1$ representing $H$, and a finite symmetric generating set $X$ for $G$. Write $\overline{g}$ for the chosen representative of $Hg$. The finitely many Schreier generatorsbelong to $H$. If $h=x_1\cdots x_k\in H$, set $t_j=\overline{x_1\cdots x_j}$, so $t_0=t_k=1$. ThenEach factor is a listed generator because $H{x}_1\cdots x_j=H{t}_{j-1}{x}_j$. Therefore $H$ is finitely generated. This proves Schreier's lemma directly.

An infinite locally finite group isIts elements are binary sequences of finite support, with coordinatewise addition. It is infinite because the unit vectors are distinct. Any finite collection of elements has its supports contained in one finite set of coordinates, and the subgroup it generates lies in the corresponding finite product of copies of $C_2$. Hence every finitely generated subgroup is finite.

Local finiteness passes to subgroups because a finitely generated subgroup of a subgroup is also one of the ambient group. It passes to quotients as well: lift finitely many group generators of a subgroup in the quotient; the subgroup generated by their lifts is finite, and its image is exactly the desired subgroup.

For local finiteness is closed under extensions, suppose $N$ and $Q=G/N$ are locally finite, and take a finitely generated subgroup $K\le G$. Its image $\overline{K}\le Q$ is finitely generated, hence finite. The map $K\to \overline{K}$ has kernel $K\cap N$, soThe finite-index result already proved makes $K\cap N$ finitely generated. Local finiteness of $N$ makes it finite, and thenSince $K$ was arbitrary, $G$ is locally finite. Thus finite generation is preserved by quotients, extensions and finite-index subgroups; local finiteness is preserved by subgroups, quotients and extensions.

The assumed group is a Tarski monster group. For any $g\ne 1$, its cyclic subgroup $⟨g⟩$ must be proper. Otherwise $G$ would be infinite cyclic, and $⟨g^2⟩$ would be a proper nontrivial infinite subgroup, contrary to the hypothesis. Hence $⟨g⟩$ has order $p$.

Choose $h\notin ⟨g⟩$, possible because $G$ is infinite. The subgroup $⟨g,h⟩$ contains $⟨g⟩$ strictly. If it were proper, it would have order $p$ and could not strictly contain the order-$p$ subgroup $⟨g⟩$. ThereforeIn particular, $G$ is a finitely generated group, generated by two elements.

Suppose $1\ne N⊲G$ is proper. The defining property makes $N$ cyclic of order $p$. Conjugation gives a homomorphismwhose kernel is the centralizer $C_G(N)$. The automorphism group has order $p-1$, so the kernel has finite index in $G$. It contains $N$, because $N$ is abelian. If this kernel were proper, it would be an order-$p$ subgroup, and its finite index would force $G$ itself to be finite. Therefore $C_G(N)=G$, so $N$ is central.

Choose $h\notin N$. Part (i) shows that $h$ has order $p$. Its cyclic subgroup meets $N$ trivially, since two order-$p$ subgroups with nontrivial intersection are equal. Since $N$ is central,This subgroup has order $p^2$ and is proper because $G$ is infinite, contradicting the assumed order of every proper nontrivial subgroup. Thus no such $N$ exists:Together with part (i), this proves that Tarski monster groups are two-generated and simple.

For the replacement condition, interpret “every element has order $p$” in the usual exponent-$p$ sense: every nonidentity element has order $p$, while the identity has order one. ConsiderEvery nonzero element has order $p$. It is infinite, but no finite collection generates it: the supports of finitely many group generators cover only finitely many coordinates, so a unit vector outside those coordinates is missing from their generated subgroup. Also $E$ is abelian, and one coordinate subgroup $C_p$ is a proper nontrivial normal subgroup. ThereforeThe stronger restriction on all proper subgroups, rather than just individual element orders, is what forces the conclusions in parts (i) and (ii).