Quotients of a 2-generated dense subgroup of a Cartesian product of infinitely many finite alternating groups

Question

A "randomly chosen" 2-generated dense subgroup $$ G \ = \ \langle a, b \rangle \ < \ {\rm A}_5 \times {\rm A}_6 \times {\rm A}_7 \times \dots $$ of the cartesian direct product of the finite simple alternating groups is "typically" free of rank 2. The "interesting" cases are those where such group $G$ is not free -- in particular those where all upper composition factors (i.e. composition factors of finite quotients) are alternating groups.

Choosing a pair of generators $(a,b)$ amounts to choosing pairs $(a_n,b_n)$ of generators for each of the direct factors ${\rm A}_n$.

Question: Assume that we choose $a_n := (1,2,3)$, and $b_n := (1,2, \dots, n)$ for $n$ odd and $b_n := (2,3, \dots, n)$ for $n$ even. Does this ensure that all upper composition factors of $G$ are alternating groups, and that none of them occurs with multiplicity greater than 1? -- And if not, which other choice of pairs $(a_n,b_n)$ of generators for the ${\rm A}_n$ serves the purpose (provided that there is such a choice)?

Answer 1

A very close variant of this group (the one in the question) was introduced and studied by B.H. Neumann (Neumann, B. H. Some remarks on infinite groups. J. Lond. Math. Soc. 12, 120-127 (1937).).

Namely, it's the same group, but restricting to the product of $A_n$ for odd; this shouldn't make much difference.

This group has a homomorphism onto $\mathbf{Z}$ which measures, for large $n$, how a given element "eventually" shifts $(1,\dots,n)$. So technically the answer is no since all prime cyclic groups occur as quotient.

This is, however, the only deviation to the expected behavior. Indeed, this group has a characteristic subgroup $W$, which is the set of elements of finite support. Modding out, what the quotient group $H$ is isomorphic to the subgroup of permutations of $\mathbf{Z}$ generated by $(123)$ and the shift $n\mapsto n+1$. This group contains the alternating group $A$ of $\mathbf{Z}$ as normal subgroup and the quotient is infinite cyclic. Since $A$ is (infinite) simple (Onofri 1929 / J.Schreier-Ulam 1934), every finite quotient of $H$ is a quotient of $H/A$, hence cyclic. Also, $W$ is just the direct sum $\bigoplus_{2n+1\ge 5}A_{2n+1}$. All this is in Neumann's paper.

Hence, if $F$ is a finite quotient of $G$, and $M$ is the image of $W$ in $F$, we deduce that $M\simeq\prod_{n\in I}A_n$ for some finite subset $I$ of $\mathbf{N}_{\ge 5}$ and $F/M$ is cyclic. In particular, the composition factors of $F$ are abelian, or alternating, and the alternating ones have multiplicity $\le 1$.

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If you allow even $n$ to exactly fit the original question, probably the conclusion is the same: the only minor modification is probably just to check that the cycles $(1,2,3)$ and $(2,3,\dots,n)$ generate $A_n$ for even $n$, which certainly holds for large $n$ and should be checkable by hand anyway (and with computer, say for $n=6,8$).

Answer 2

This question appears in a related form in an article of L. Pyber (p. 252): Old groups can learn new tricks

More precisely he asked whether the product of all alternating groups $Alt(n)$, where $n \geq 5$ is an odd integer, can be realized as the profinite completion of a finitely generated group.

As Yves pointed out, the groups of B.H. Neumann are very close to answer this. However, it turns out that it is difficult to get rid of the finite cyclic quotients. This was managed by M. Kassabov and N. Nikolov by constructing a 22-generated group $G$ with $\widehat{G} \cong P := \prod \limits_{n=5}^{\infty} Alt(n)$ (Proposition 3.4 in Cartesian producs as profinite completions).

Thus without the restriction that the subgroup of $G < \prod \limits_{n=5}^{\infty} Alt(n)$ is 2-generated, the question of Stefan is solved. In fact the question whether 2-generation is possible in this case is a special case of a question of M. Kassabov and N. Nikolov (Question 1.6 in Cartesian producs as profinite completions) which asks whether the minimal cardinality of a topological generating set of a product $Q$ of finite simple groups can be realized as the cardinality of a generating set of a discrete group $G$ with $\widehat{G} \cong Q$ (under the assumption that $Q$ is the profinite completion of a finitely generated group). As far as I know the only cases where this is known (and true) are provided by S.Kionke and myself for certain products of alternating groups and finite special linear groups (Theorem 3.11 and Theorem 3.13 in From telescopes to frames and simple groups).