Let $K<H<G$ be a chain of subgroups. Let $R \subset G$ be a system of representatives for $G/H$ and let $S \subset H$ be a system of representatives for $H/K$.

I have to show the following:

(a) Show that the map $R \times S \rightarrow RS$ given by $(r,s) \mapsto rs$ is a bijection.

(b) Show that $RS = \{rs | r \in R, s \in S\}$ is a system of representatives for $G/K$ and conclude that $[G:K] = [G:H][H:K]$.

I am completely lost on how to prove both (a) and (b) and would appreciate any help.

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    $\begingroup$ For (a), you need to show that the map is an injection and a surjection. To show that it is an injection, suppose that $r_1,r_2 \in R$ and $s_1,s_2 \in S$, and $r_1 s_1 = r_2 s_2$. This means that $r_2^{-1} r_1 = s_2 s_1^{-1}$. Therefore, $r_2^{-1}r_1 \in H$, so $r_1 H = r_2 H$, in other words $r_1$ and $r_2$ are in the same coset of $H$. But $R$ is a system of representatives of $G/H$, so... $\endgroup$ – Bungo Oct 14 '15 at 6:16
  • $\begingroup$ @Bungo, how did you get that $r_2^{-1}r_1\in H$? $\endgroup$ – Bihc Oct 15 '15 at 0:42
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    $\begingroup$ @Bihc: $r_2^{-1}r_1 = s_2 s_1^{-1}$, and $s_1$ and $s_2$ are in $S$, which is a subset of $H$. Since $H$ is a subgroup, $s_2 s_1^{-1}$ is also in $H$. $\endgroup$ – Bungo Oct 15 '15 at 0:44
  • $\begingroup$ @Bungo, any ideas for surjectivity? $\endgroup$ – Bihc Oct 15 '15 at 1:54
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    $\begingroup$ @Bihc: Sure, an arbitrary element of $RS$ is of the form $rs$, where $r \in R$ and $s \in S$. So the element $(r,s) \in R \times S$ is mapped to $rs$. $\endgroup$ – Bungo Oct 15 '15 at 1:56

May as well write an answer since part (a) has been done in the comments and (b) is a natural consequence of (a).

Let's give the map in part (a) a name: $\phi : R \times S \to RS$, where $\phi(r,s) = rs$.

We first show that $\phi$ is an injection. Suppose $r_1 s_1 = r_2 s_2$. Then $r_2^{-1} r_1 = s_2 s_1^{-1}$. Since $s_1$ and $s_2$ are in $S$, which is a subset of $H$, and $H$ is a subgroup, it follows that $s_2 s_1^{-1} \in H$, so $r_2^{-1} r_1 \in H$. This means that $r_1 H = r_2H$, in other words $r_1$ and $r_2$ are in the same coset of $H$. Since $R$ contains exactly one representative from each coset of $H$, we must have $r_1 = r_2$. But then $r_2^{-1} r_1 = 1$, the identity element, and this means that $s_2 s_1^{-1}$ is also the identity, so $s_1 = s_2$.

Now we show that $\phi$ is a surjection. This is immediate from the definition, because an arbitrary element of $RS$ is of the form $rs$, where $r \in R$ and $s \in S$, and therefore $rs$ is the image under $\phi$ of the element $(r,s) \in R \times S$.

We conclude that $\phi$ is a bijection.

For part (b), define $\psi : R \times S \to G/K$ by $\psi(r,s) = rsK$.

We show that $\psi$ is an injection. Suppose that $r_1 s_1 K = r_2 s_2 K$. Since $s_1 K \subset H$ and $s_2 K \subset H$, it follows that $r_1 s_1 K \subset r_1 H$ and $r_2 s_2 K \subset r_2 H$. Therefore, $r_1 H$ and $r_2 H$ are not disjoint, so they must be equal: $r_1 H = r_2 H$. This implies that $r_1 = r_2$ by definition of $R$. Let us set $r = r_1 = r_2$. Then we have $rs_1 K = rs_2 K$, so $rs_1 \in rs_2 K$, and therefore $s_1 \in s_2 K$, which means that $s_1 K = s_2 K$. By definition of $S$, this means that $s_1 = s_2$.

We show that $\psi$ is a surjection. If $g$ is an arbitrary element of $G$, then it lies in some element of $G/H$, say $g \in rH$. Then $r^{-1}g \in H$, so $r^{-1}g$ lies in some element of $H/K$, say $r^{-1}g \in sK$. Then $g \in rsK$, so $gK = rsK = \psi(r,s)$.

Finally, note that the statement "$RS$ is a system of representatives of $G/K$" is equivalent to the statement "there is a bijection from $RS$ to $G/K$ which maps $rs \to rsK$". But this is now clear: $\psi \circ \phi^{-1} : RS \to G/K$ is the desired bijection.

Edit to address the question raised in the comment and in the edited version of the question.

We want to conclude that $[G:K] = [G:H][H:K]$.

Note that this follows immediately from what we have already done.

Let's assume that $[G:K]$ is finite. Then, by definition, $[G:K]$ is the number of elements in $G/K$. We have shown that there is a bijection between $G/K$ and $RS$, so $[G:K]$ is also equal to the number of elements in $RS$, which we may denote as $|RS|$. Similarly, $[G:H] = |G/H| = |R|$ and $[H:K] = |H/K| = |S|$. So the equation $[G:K] = [G:H][H:K]$ is the same as $|RS| = |R| |S|$. This in turn is true because we have a bijection between $RS$ and $R \times S$, and the size of $RS$ is $|RS|$, and the size of $R \times S$ is $|R| |S|$.

If $[G:K]$ is infinite, then one or both of $[G:H]$ or $[H:K]$ must also be infinite, again because of the bijection between $RS$ and $R \times S$. So the equation holds (trivially) in that case as well: both sides are $\infty$.

  • $\begingroup$ Thanks very much for a very clear answer. How do I now show that $[G:K] =[G:H][H:K]$? Intuively, I understand how it is the case for finite groups but generalizing I find unclear. $\endgroup$ – letsmakemuffinstogether Oct 15 '15 at 18:45
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    $\begingroup$ @letsmakemuffinstogether: Actually, since it's related to this question (indeed, I see you edited the question to add it), I'll go ahead and answer it in an edit. $\endgroup$ – Bungo Oct 15 '15 at 18:52
  • $\begingroup$ Thanks for your time, your edit is clear. $\endgroup$ – letsmakemuffinstogether Oct 15 '15 at 19:05
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    $\begingroup$ @letsmakemuffinstogether: That's right, $r_2^{-1}r_1 \in H$ if and only if $r_2^{-1}r_1 H = H$ if and only if $r_1 H = r_2 H$ if and only if $r_1 \in r_2 H$ if and only if $r_2 \in r_1 H$. These are all equivalent. They all mean that $r_1$ and $r_2$ are in the same coset of $H$. $\endgroup$ – Bungo Oct 17 '15 at 17:56
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    $\begingroup$ @letsmakemuffinstogether: It's the same argument as in my previous comment. Since $g \in rsK$, it means that $g$ is in the coset $rsK$, so the coset containing $g$, which is $gK$, is the same as the coset $rsK$. $\endgroup$ – Bungo Oct 17 '15 at 19:30

Bungo handed me a) on a platter. But here are my thoughts for b):

Since we don't know if any of the groups are finite, let $R=\{g_1,g_2,...\}$ and $S=\{h_1,h_2,...\}$ where $$\bigcup_{r\in R}rH=G,\quad\text{and}\quad\bigcup_{s\in S}sK=H$$ by definition of $R,S$ being sets of representatives. Now consider any $g\in G$. Then $g=r_{\alpha}H$ for some $\alpha\in\mathbb{Z}$, hence there is some $h\in H$ s.t. $g=r_\alpha h$. Now consider $h\in H$. Then $h=s_\beta K$, some $\beta\in\mathbb{Z}$, so $\exists k\in K$ s.t. $h=s_\beta k$. Then we can simply compute $$g=r_\alpha h=r_\alpha s_\beta k$$ Hence $g\in r_\alpha s_\beta K$ where $r_\alpha s_\beta =g'$ for some $g'\in G$. Since $g$ was arbitrary we can conclude that every $RS$ contains a representative for every $g\in G$.

  • $\begingroup$ Your use of $\alpha \in \mathbb Z$ suggests that you assume there are only countably many cosets. This need not be the case. For exaxmple, if $G = \mathbb R$ under addition and $H = \mathbb Q$, then there are uncountably many cosets of $\mathbb Q$. This is a relatively minor detail, though. The larger issue is that you have only shown that every coset in $G/K$ has a representative of the form $rs$. You have not shown that each coset contains exactly one representative of that form. $\endgroup$ – Bungo Oct 15 '15 at 4:47
  • $\begingroup$ @Bungo Yeah that's a good point about countability, I didn't even consider that. I like your approach, very smooth. $\endgroup$ – Bihc Oct 15 '15 at 6:11

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