Let $W:=\prod_{i\in \omega} F_i$ be the (external) unrestricted direct product and $U:=\prod_{i\in \omega}^w F_i$ be the (external) restricted direct product of finite groups $F_i$ such that $|F_{i}|<|F_{i+1}|$ for every $i\in\omega$. What can be said about $W/U$? Is $W/U$ residually finite (while $W$ is residually finite), for example?
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2$\begingroup$ Take $F_i=\mathbf{Z}/p_i\mathbf{Z}$, where $p_i$ is prime and $p_i\to\infty$. Then your quotient contains $\mathbf{Q}$ as subgroup, hence is not residually finite. $\endgroup$– YCorCommented Aug 3, 2021 at 22:13
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$\begingroup$ If the $F_i$ have unbounded exponent then the near product ($\prod F_i/\bigoplus F_i$) is also not residually finite. I even think it's not an iff condition: fix an odd prime $p$ and take the Heisenberg group of order $p^{2n+1}$ (central product of $n$ copies of the group of order $p^3$ of exponent $p$): the near product might fail to be residually finite. $\endgroup$– YCorCommented Aug 4, 2021 at 10:15
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$\begingroup$ The condition that $|F_{i}|<|F_{i+1}|$ is redundant. If $\prod_{i\in\omega}G_{i}$ is a product of finite groups and infinitely many of these groups are non-trivial, then there is an increasing sequence $(n_{k})_{k\in\omega}$ with $n_{0}=0$ such that if $F_{k}=G_{n_{k}}\times\dots\times G_{n_{k+1}-1}$, then $|F_{k}|<|F_{k+1}|$ for all $k$, and here $\prod_{k\in\omega}F_{k}/\text{Fin}\simeq\prod_{i\in\omega}G_{i}/\text{Fin}$. $\endgroup$– Joseph Van NameCommented Aug 18, 2021 at 12:44
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Let $F_i$, $i\ge 1$, be a sequence of finite groups such that for every prime $p$ the orders of all but finitely many groups $F_i$ are not divisible by $p$. Then your group $W/U$ has no homomorphism onto a nontrivial finite group. Indeed if $G$ is a finite nontrival homomorphic image and $\phi: W/U\to G$ is a homomorphism, $n=|G|$, $1\ne a\in G$, $b\in W/U, \phi(b)=a$ then $b=c^n$ for some $c\in W/U$, therefore $a=\phi(c)^n=1$ in $G$, a contradiction. In particular $W/U$ is not residually finite. I used the fact that if a prime $p$ does not divide the order of a finite group $F$ then every element of $F$ has a root of degree $p$. Thus every element of $W$ has a root of degree $p$ modulo $U$.
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$\begingroup$ Many thanks for the comment and the answer. So there seems to be some possibilities to be residually finite. $\endgroup$– IGTCommented Aug 4, 2021 at 7:49
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$\begingroup$ I did not add any comment.Of course, sometimes $W/U$ is residually finite. For example if all $F_i$ are Abelian of exponent $2$ then $W/U$ is Abelian of exponent 2 and so is residually finite. $\endgroup$– markvsCommented Aug 4, 2021 at 8:21
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$\begingroup$ (You need to ask $|F_i|>1$ — actually it's enough to have just one prime divisor of the order tending to infinity). This answer actually follows from my comment, since this contains $Z/p_iZ$ for some primes $p_i\to\infty$, and hence the "near product" contains a copy of $Q$ (actually, of $R$). $\endgroup$– YCorCommented Aug 4, 2021 at 10:02
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$\begingroup$ One step further: $W/U$ may have a $p$-subgroup $T/U$ for some prime $p$. So $T/U \cap Q/U=1$ where $Q/U$ is the isomorphic copy of $\mathbb{Q}$ in $W/U$. Is $T/U$ residually finite in this case? $\endgroup$– IGTCommented Aug 6, 2021 at 22:09
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$\begingroup$ If $T/U$ is infinite, it may be isomorphic to the Prüfer group $C_{p^\infty}$ which is not residually finite. $\endgroup$– markvsCommented Aug 6, 2021 at 23:18