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For a group $G$ and its subgroup $H$, the index of $H$ relative to $G$, denoted by $[G:H]$, is the cardinality of the set $\{ gH \mid g \in G \}$.

It is known that $|G| = [G:H]|H|$ even if all cardinalities in question are infinite. However, what about the the identity $[G:H] = [G:K][K:H]$ for subgroups $H \subseteq K$ of $G$? Is it true? If so, how to prove it?

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    $\begingroup$ The same proof technique should do. Make a bijection between $G/H$ and $G/K\times K/H$. $\endgroup$ – Algeboy Feb 4 at 10:31
  • $\begingroup$ @Algeboy Well, constructing a bijection is a standard way to prove an equinumerousity of two sets. The question is, which bijection can be defined between these two sets? :) $\endgroup$ – Jxt921 Feb 4 at 10:47
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Lemma. Let $X$ be a set and $\sim$ an equivalence relation on it. Assume that for any $x,y\in X$ we have a bijection $f_{x,y}:[x]\to[y]$. Then for any $x\in X$ there exists a bijection $[x]\times X/\sim$ $\to X$.

Proof. So the idea here is that if all equivalence classes are equinumerous then every element is uniquely determined by its equivalence class and the corresponding position in the fixed equivalence class. Formally define function

$$F:[x]\times X/\sim \; \to X$$ $$F(a, [b])=f_{x,b}(a)$$

Now under the assumption that $f_{x,y}=f_{x',y'}$ if $[x]=[x']$ and $[y]=[y']$, the function is a bijection; I leave that as an exercise. Note that the assumption can be easily enforced by replacing "broken" bijections by a fixed one per pair of equivalence classes. $\Box$


Now take $G/H$ and define $\sim$ on it: $gH\sim g'H$ iff $gK=g'K$. Note that for any $g$ we have a bijection $K/H\to [gH]$ given by $kH\mapsto gkH$. So our lemma applies, resulting in a bijection between $G/H$ and $K/H\times (G/H)/\sim\,.$ All that is left is to see that $(G/H)/\sim$ is equinumerous with $G/K$. Can you complete the proof?

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