Group algebra of a subgroup

Question

I already asked this question on MathOverflow, but it was suggested that I ask it here. Below follows my question.

I think what I want to do is not very hard, but I am just not seeing it.

Let $K$ be a subgroup of a finite group $H$, and let $A$ be a commutative ring with $1$. Consider the group algebras $A[K]\subset A[H]$. I want to show that $A[H]$ is free as a module over $A[K]$.

My proof so far

I think that the quotient $A[H]/A[K]$ is finite (if this is true, how can I show this?). Then we can take the coset representatives $x_1,\dots, x_n\in A[H]$, where $n$ is the cardinality of the quotient. I think that the $x_i$ form a basis for $A[H]$ over $A[K]$ (again, how can I see this?).

Any help is appreciated!

Answer 1

Hint: Let $h_1,\dots,h_n$ be representatives of the right cosets: $H=\sqcup_i Kh_i.$ Check that the map $$(A[K])^n\to A[H],\quad(x_1,\dots,x_n)\mapsto\sum_ix_ih_i$$ is an isomorphism of (left) $A[K]$-modules.