# If G is totally residually cyclic, does G contend the majority of one subgroup of each limited index?

Update : Steve mentions in remarks that several straight items of residually cyclic teams will certainly be counterexamples. Without a doubt, I need to have detected this : every finitely created abelian team is residually cyclic! I need to have inquired about **totally ** residually cyclic teams, ie teams $G$ such that for any kind of limited part $X\subseteq G\smallsetminus 1$, there is a homomorphism from $G$ to a cyclic team that does not eliminate any kind of components of $X$. I believe the inquiry is currently fairly very easy, though I have not had an opportunity to think of it yet.

This inquiry is encouraged by this recent question, which requested for a characterisation of teams with at the majority of one subgroup of each limited index. Arturo Magidin is solution revealed that every such limited team is cyclic, yet the questioner, Louis Burkill, included remarks that he is actually curious about the boundless instance.

In my response to that inquiry, I say that a *finitely created * team contends the majority of one subgroup of each limited index if and also just if its approved residually limited ratio, $R(G)$, is cyclic. The 'finitely created' theory is essential - or else the additive team of the rationals gives a counterexample.

In the basic instance, one can lower from the instance of $G$ to $R(G)$ as in the past (this removes pathological instances like boundless straightforward teams), and also the very same argument reveals that if $R(G)$ contends the majority of one subgroup of each limited index after that $R(G)$ is *totally * residually cyclic, specifically abelian. Yet the reverse is unclear to me. Therefore the inquiry in the title, which I'll state below, with some added theories that need to show where the trouble exists.

If $G$ is a definitely created,

totallyresidually cyclic (specifically, abelian) team, must $G$ contend the majority of one subgroup of each limited index?

**The probabilities protest us. Also "fully" is not nearly enough. **
With the "fully" theory, possibly it is very easy : If $G$ has 2 subgroups $H,K$ of limited index $n$, after that their junction has limited index. Take $X$ to be a set of (non - identification) coset representatives of $G/H \cap K$, after that there is a ratio where no component non - identification component of $G/H \cap K$ is sent out to the identification ~~ (and also WLOG, G/H \cap K is that ratio) ~~, yet given that $G$ is totally residually cyclic, that ratio is cyclic, and also ~~both subgroups have to equal by the latticework homomorphism theory ~~.

~~To put it simply, you appear to have actually specifically addressed your very own inquiry! ~~

At the very least K4 is no more a counterexample.

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