Exercise 9.2.4

Proof. Let $\{b_1\cdots ,b_d\}$ a complete system of representatives of left cosets of $A$ in $B$, where $d=(B:A)$. Then

If $\mathit{cB},c\in G$, is any left coset of $B$ in $G$, then

Indeed,

$\text{∙}$

$b_{i}A\subset B$, thus $c{b}_{i}A\subset \mathit{cB}$, $i=1,\dots ,d$, therefore ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{1\le i\le d}c{b}_{i}A\subset \mathit{cB}$.

$\text{∙}$

If $g\in \mathit{cB}$, then $g=\mathit{ch},h\in B$, and $h\in b_{i}A$ for some $i,1\le i\le d$, so $h=b_{i}a,a\in A$, hence $g=c{b}_{i}A\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{1\le i\le d}c{b}_{i}A$ . Therefore $\mathit{cB}\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{1\le i\le d}c{b}_{i}A$. $$\mathit{cB}=\underset{1\le i\le d}{\bigcup }c{b}_{i}A.$$

$\text{∙}$

The union is a disjoint union: if $g\in c{b}_{i}A$ and $g\in c{b}_{j}A$, then $c^{-1}g\in b_{i}A\cap b_{j}A$, which is possible only if $i=j$. Thus $i\ne j\Rightarrow c{b}_{i}A\cap c{b}_{j}A=\varnothing$.

Conclusion: every left coset of $B$ in $G$ is the disjoint union of $d=(B:A)$ left cosets of $A$ in $G$. □