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Let $K\leq H\leq G$ (not necessarily finite groups).

Why do we have $[G:K]=[G:H]\cdot [H:K]$? I can't figure out a proof in the setting of possibly infinite groups and non-normal subgroups.

hululu
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JACK3D
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    Possible duplicate? http://math.stackexchange.com/questions/730728/if-k-leq-h-leq-g-show-that-gk-ghhk

    http://math.stackexchange.com/questions/1169108/let-g-be-a-probably-infinite-group-and-h-leq-k-leq-g-gh-infty-r

     – hululu Jul 24 '16 at 10:50
  • You could also try proving the map $G/K\to G/H$ is a $|H/K|$-to-$1$ map. – anon Jul 24 '16 at 11:33

1 Answers1

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If we can enumerate $H/K$ as $h_iK$, $i\in I$, and $G/H$ as $g_jH$, $j\in J$, then $g_jh_iK$, $(i,j)\in I\times J$, happens to enumerate $G/K$:

To show that we reach all cosets, let $gK$ be any element of $G/K$. Then $g\in g_jH$ for some $j\in J$ and from $K\le H$, we see $gK\subseteq g_jH$, so $g_j^{-1}gK\subseteq H$. Now $g_j^{-1}g\in h_i K$ for some $i\in I$. That makes $g_j^{-j}gK=h_iK$ and so $gK=g_jh_i K$.

To show that no coset is obtained repeatedly, assume $g_jh_i K=g_{j'}h_{i'}K$. Then $g_j H$ and $g_{j'}H$ intersect, which means $g_j=g_{j'}$. But then $h_iK=h_{i'}K$, so $h_i=h_{i'}$.

Hagen von Eitzen
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  • "...from $K\le H$, we see $gK\subseteq g_jH$, so $g_j^{-1}gK\subseteq H$...", this part is unnecessary right? Already have $g\in g_jH$, which leads directly to $ g_j^{-1}g \in H$ and hence $g_j^{-1}g\in h_i K$ for some $i\in I$ because $H$ is partitioned by $K$. – Geralt of Rivia Oct 18 '17 at 16:22
  • May I delete that part? – Geralt of Rivia Oct 18 '17 at 16:24