Group algebra is a free module of finite rank over any group sub-algebra

Question

Let $G$ be a finite group and let $H$ be a subgroup. Let $\mathbb CG$ and $\mathbb CH$ be the corresponding group algebras. Clearly, the injective group homomorphism $H\to G$ defines a $\mathbb CH$-module structure on $\mathbb CG$. The question is if $\mathbb CG$ is a free left and right module of finite rank over $\mathbb CH$. The claim is true in the case $H:=\{1_G\}$ but is it true for any subgroup of $G$? Is any set $\{g_1, ..., g_k\}$ of representatives of the left cosets $G/H$ a basis of $\mathbb CG$ over $\mathbb CH$?

Answer 1

Given a set of right coset representatives $\{g_1, \dots, g_k \}$, you need to check that the map $$(\mathbb{C}H)^k \to \mathbb{C}G$$ $$(h_1, \dots, h_k) \to \sum_{i=1}^k h_i g_i$$

is an isomorphism of left $\mathbb{C}H$-modules. After showing that this is indeed a homomorphism, injectivity is easy, using that each $g_i$ lies in a different coset. For surjectivity, you need to find a preimage of an arbitrary $g \in G$ and the rest follows. For that, take the $g_j$ that represents its right coset and the appropriate $h$ such that $g=hg_j$.