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Let $G$ be a group. Let $H$ be a subgroup of $G$ such that $(G : H) \lt \infty$. Then there exists a sequence of elements $a_1,\cdots, a_n$ such that $G = \bigcup_{i=1}^n a_iH$ is a disjoint union. Clearly $\sum_{i=1}^n 1/(G : H) = 1$.

Now I would like to generalize this formula. Suppose there exist a sequence of elements $a_1,\cdots, a_n$ and a sequence of subgroups $H_1,\cdots, H_n$ such that $G = \bigcup_{i=1}^n a_iH_i$ is a disjoint union and $(G : H_i) \lt \infty$ for all $i$.

My question is:

$\sum_{i=1}^n 1/(G : H_i) = 1$?

Note that $H_1 = \cdots = H_n$ does not generally holds even when $(G : H_1) = \cdots = (G : H_n)$. See Finite partition of a group by left cosets of subgroups

I came up with this problem when trying to explicitly construct Haar measure on a profinite group: Explicit construction of Haar measure on a profinite group

Here is a train of my thoughts. Let $G$ be a profinite group.

It is known that every neighborhood of the identity element of $G$ contains an open compact subgroup. Since $G$ is a compact Hausdorff group, there exists a Haar measure $\mu$ on $G$ such that $\mu(G) = 1$. Let $H$ be an open compact subgroup of $G$. Since the set of left cosets $G/H$ is compact and discrete with its quotient topology, $(G : H) \lt \infty$. Hence there exists a finite sequence of elements $a_1,\cdots, a_n$ such that $G = \bigcup_{i=1}^n a_iH$ is a disjoint union. Then $1 = \mu(G) = \sum_{i=1}^n \mu(a_iH) = \sum_{i=1}^n \mu(H) = n\mu(H)$. Hence $\mu(H) = 1/n = 1/(G : H)$.

Now suppose there exist a sequence of elements $a_1,\cdots, a_n$ and a sequence of open compact subgroups $H_1,\cdots, H_n$ such that $G = \bigcup_{i=1}^n a_iH_i$ is a disjoint union. Then $1 = \mu(G) = \sum_{i=1}^n \mu(a_iH_i) = \sum_{i=1}^n \mu(H_i) = \sum_{i=1}^n 1/(G : H_i)$.

I wondered if this formula holds on general abstract groups.

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  • What, precisely, is your question? Are you asking if the second paragraph is true or for a proof or ...? – Michael Burr Aug 30 '15 at 21:25
  • @MichaelBurr I have edited the question to make it clearer. – JHW Aug 30 '15 at 21:27
  • You could take the intersection of all the $H_i$'s and that will give a finite index subgroup. Then, if you apply the theorem to each $H_i$, you should get your answer. – Michael Burr Aug 30 '15 at 21:29
  • @MichaelBurr Would you please make it as an answer so that I could accept it and the thread would be completed. – JHW Aug 30 '15 at 21:31

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Let $H=H_1\cap \cdots\cap H_n$. By finite index subgroup, the intersection of the $H_i$'s is a finite index subgroup.

Then, $\sum_{i=1}^m\frac{1}{[G:H]}=1.$ Now, however, we can write each $H_i$ as a union of $k_i$ cosets of $H$, so that

$$ 1=\sum_{i=1}^m\frac{1}{[G:H]}=\sum_{i=1}^n\sum_{j=1}^{k_i}\frac{1}{[G:H]}=\sum_{i=1}^n\sum_{j=1}^{k_i}\frac{1}{[G:H_i][H_i:H]}=\sum_{i=1}^n\frac{1}{[G:H_i]}\sum_{j=1}^{k_i}\frac{1}{[H_i:H]}=\sum_{i=1}^n\frac{1}{[G:H_i]} $$ by applying the original statement to each $H_i$.

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