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Suppose I have a group $G$ and a commutative subgroup $H$ and another subgroup $K\subseteq H$. So I have $K\leq H\leq G$. Moreover, suppose I know that the index $\vert G:H\vert$ is finite. I have to prove or disprove that $\vert G: K\vert $ is finite.

My attempt: I think it is true. So I tried proving the statement. By Lagrange theorem we have $$\vert G:K\vert = \vert G:H\vert \vert H:K\vert.$$ Thus

$$\frac{\vert G:K\vert}{\vert H:K\vert} $$ is finite. Which implies that $\vert H:K\vert$ is finite.

I am very iffy about this proof. Since some things could be infinite so it feels strange to me... Can someone help me understand this better or push me into the right direction?

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    $\begingroup$ Lagrange typically assumes everything is already finite. Unless you're using a cardinal-specific version, don't see how you can argue this way. $\endgroup$
    – Randall
    Commented Oct 21 at 12:46
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    $\begingroup$ Also hint: it's very false. Easy examples exist where $K$ is trivial. $\endgroup$
    – Randall
    Commented Oct 21 at 12:52
  • $\begingroup$ If any of the quantities $[G:K]$ and $[H:K]$ is infinite, then you cannot do "division" as you are doing. $\endgroup$
    – Arturo Magidin
    Commented Oct 21 at 15:28

2 Answers 2

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Let $H=G$ be an infinite group. (You stipulate that $H$ must be abelian; that is not necessary here.) Then $[G:H]=1$. Let $K$ be the trivial subgroup.

The problem with your attempt is that you cannot divide things that are infinite.

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    $\begingroup$ Best example. Also teaches well that beginners should always look at the edge cases. No need to come up with complicated examples before doing that. $\endgroup$
    – Martin Brandenburg
    Commented Oct 21 at 21:25
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    $\begingroup$ I think claiming that the "$H$ abelian" assumption is not necessary is a bit misleading, since it's part of the premise. Showing a counterexample to a weaker statement (eg the same question without the abelian assumption) does not necessarily prove the stronger statement. In this case the answer still works, of course, since we can readily find an abelian infinite group - I just find the phrasing confusing. $\endgroup$
    – MartianInvader
    Commented Oct 21 at 23:59
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It is false. Counterexamples: (1) Take $G=\mathbf{Z}$, $H= n\mathbf{Z}$, and $K =(0)$ for some positive integer $n$. (2) Take $G = \mathbf{Z}[1,(-1+√5)2]$, $H = \mathbf{Z}[1,√5]$ and $K = \mathbf{Z}$. The problem with your solution is that $\mid G : K \mid$ and $\mid H : K \mid$ may both be infinite (as in the examples above).

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