I'm translating a paper from the 70's where the author cites László Fuchs' Abelian groups (the entire book, no page number). As you can imagine, given the size and differences in editions of Fuchs' book, I am having trouble connecting the dots.

Here is the situation: $G$ is an infinite, abelian, and locally finite group. $H$ is an infinite subgroup of $G$ which is of finite index in $G$. From this, the step in a proof I am reading continues this way

... we deduce the socle of $G$ is of finite length and that $G$ is an Artinian group.

The justification of the block above is what I am asking about.

In the newest edition (2015) of Fuchs' Abelian groups, I feel think the relevant theorem is this:

Theorem 5.3 (Prufer, Kurosh, Yahya)


(i) $A$ is finitely cogenerated

(ii) $A$ is an essential extension of a finite group

(iii) $A$ is torsion of finite rank

(iv) $A$ is a direct sum of a finite number of cocyclic groups

(v) the subgroups of $A$ satisfy the minimum condition

[later] Obeserve that (ii) is equivalent to the finiteness of the socle in a torsion group.

This is the very first theorem in the section, preceded only by the definition of finite cogeneration.

The local finiteness obviously makes $G$ torsion, so I can see why both conclusions are linked... but how do you use the fact that $|G:H|$ is finite to prove one of these conditions?

I feel like I'm overlooking some connection between $G/H$ and $G$, possibly about the socles. In general module theory, there usually isn't a connection between the two, but perhaps since $G$ is locally finite there is a connection in my blind spot.

I've decided the original source is in order:

enter image description here

$A$ is a right self-injective ring, and $G$ is, as proven in an earlier step, at least a locally finite group. The $G_1$ and $H_1$ in this snippet are the $G$ and $H$ I was referring to in my original description. The citation (7) refers to Fuchs' Abelian groups. The theorem of Faith has to do with the injectivity of a free $A[H_1]$ module. Basically it allows you to conclude that $A[G_1]$ is a direct sum of finitely many copies of $A[H_1]$, and this means $|G_1:H_1|$ is finite.

Perhaps the theorem that's needed (which did not come out in my description above) is that if $G$ is a locally finite, infinite abelian group whose infinite subgroups are all of finite index, $G$ is Artinian? If this is the case, then a reference to that result would be an acceptable solution to this problem. (Hopefully in Fuchs, but elsewhere would be fine.)

  • $\begingroup$ Without any other assumptions on $H$ you can always take $G=H,$ so $H$ is irrelevant. An infinite product of $C_2$’s is a counterexample (non Artinian) unless I’m missing something. $\endgroup$ – Dap Jan 2 '18 at 17:43
  • $\begingroup$ @Dap You're right... I thought i presented the problem faithfully but I will try to double-check. $\endgroup$ – rschwieb Jan 2 '18 at 18:14
  • $\begingroup$ @Dap I've added some of the original source material. Perhaps there are stronger properties available that I overlooked. This is the only step of the proof that I don't get, and I'd like to figure out for sure if there is a gap here or not. $\endgroup$ – rschwieb Jan 2 '18 at 18:28
  • $\begingroup$ @Dap I think I might have seen what I accidentally occluded in my original description. If you have a locally finite, infinite abelian group, and all infinite subgroups have finite index, does that imply the five conditions in the theorem? $\endgroup$ – rschwieb Jan 2 '18 at 18:36
  • $\begingroup$ @rschwieb: Let P be the property that all infinite subgroups have finite index. If G has Property P, then so does any infinite subgroup of G. But no infinite, locally finite, semisimple group has P. Thus if G has property P, its socle must be finite. Hence (ii) holds for G, since G is an essential extension of its socle. $\endgroup$ – Keith Kearnes Jan 7 '18 at 9:44

This is a problem about the class $\mathcal A$ of infinite, abelian, locally finite groups.

Let $P$ be the property that all infinite subgroups have finite index. If $G\in\mathcal A$ has Property $P$, then so does any infinite subgroup of $G$. But no infinite, locally finite, semisimple group has $P$, so if $G\in\mathcal A$ has $P$, then the socle of $G$ must be finite. This shows that (ii) holds for $G$, since any $G\in\mathcal A$ is an essential extension of its socle.

  • $\begingroup$ Thanks... all the pieces suddenly fall into place! I should have stared at that item some more... $\endgroup$ – rschwieb Jan 8 '18 at 11:51

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