Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.
So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle?
Reference: Fraleigh p. 103 Question 10.35 in A First Course in Abstract Algebra
Best Answer
In fact, the result holds even if you do not assume that we are dealing with finite index, provided you take the equality to be an equality of cardinalities. Just follow the idea Pete Clark suggests, but without assuming that the number of cosets is finite in either case. And remember that the number of distinct cosets of a subgroup $B$ in a group $A$ is equal, by definition, to the index of $B$ in $A$, $[A:B]$.
As to whether the $x_i$ are "coset representatives for $H$", remember that for a collection of elements to be "a set of coset representatives" we require that if $x_iH = x_jH$, then $i=j$, and that for every $g\in G$ there exists $i$ such that $gH = x_iH$ (thinking left cosets; right cosets work the same way). So $x_1,\ldots,x_n$ will not generally be "a coset representative" nor a set of coset representatives, but that is immaterial. Simply show that if $\{x_i\}_{i\in I}$ is a set of coset representatives for $K$ in $G$ (meaning, (i) for every $g\in G$ there exists $i\in I$ such that $gK = x_iK$; and (ii) for all $i,i'\in I$, if $x_iK = x_{i'}K$ then $i=i'$), and $\{y_j\}_{j\in J}$ is a set of coset representatives for $H$ in $K$ (meaning, (1) for every $k\in K$ there exists $j\in J$ such that $kH = y_jH$; and (2) for all $j,j'\in J$, if $y_jH = y_{j'}H$, then $j=j'$); then it follows that $\{x_iy_j\}_{(i,j)\in I\times J}$ is a set of coset representatives for $H$ in $G$ (meaning, you need to prove that (I) for all $g\in G$ there exists $(i,j)\in I\times J$ such that $gH = (x_iy_j)H$; and (II) for all $(i,j),(i',j')\in I\times J$, if $(x_iy_j)H = (x_{i'}y_{j'})H$, then $(i,j)=(i',j')$).
A fact that will no doubt be useful is to remember that for any group $A$ and any subgroup $B$ of $A$, $cB = dB$ if and only if $cB\cap dB\neq\emptyset$.