Quotients of a 2-generated dense subgroup of a Cartesian product of infinitely many finite alternating groups 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)?
 A: 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 $\le 1$.

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 $n=6,8$).
