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This isn't a duplicate. I tried kb's answer and Answerer 1 but I'm still confounded. I like $\frac {\left| G\right| } {\left| H\right| }$ better than $[G:H]$ hence I write it as a fraction.

Suppose $K \le H \le group \, G$, $\frac{|H|}{|K|}$ and $\frac{|G|}{|H|}$ are both fini te. Prove $\frac{|G|}{|K|} = \frac{|G|}{|H|}\frac{|H|}{|K|}$ is finite.

Proof : Let $\{a_iH : 1 \le i \le r \}$ be the set of distinct left cosets of H in G and
$\{b_jK : 1 \le i \le j \le s \}$ be the collection of distinct left cosets of K in H.
Show $\{(a_ib_j)K : 1 \le i \le r, 1 \le i \le j \le s \}$ is the collection of distinct left cosets of K in G.

(1.) I'm completely confounded how the three sentences after Proof: overhead relate at all to the Theorem ? I can follow the proof underneath but I can't flesh out what anything means.

(2.) How do you envisage and envision to 'let' all these things? They all look magical.

(3.) What's the intuition? enter image description here

(4.) Why is the red true? How do you induce $i = p$?

(5.) Why are Proofwiki's Proof 2 and https://math.stackexchange.com/a/22570/53934 a lot easier? Are they just simplifying fractions? Then why fret about overhead and Proofwiki's Proof 1?

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1 Answer 1

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The problem is that "$[G:H]$" and "$|G|/|H|$" do not denote the same thing when the groups in question are not finite. If they are infinite groups, the index $[G:H]$ may still be finite but "$|G|/|H|$" is not defined. So your 'change of notation' is not really legitimate here. The reason this proof is more complicated is precisely because it is dealing with the infinite case. (Note that the proofwiki proof begins "Assume $G$ is finite" whereas the undoctored version of the question here merely assumes $[H:K]$ and $[G:H]$ are finite.)

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