Free module over a group transfers to a free module over a subgroup

Question

Let $H$ be a subgroup of $G$.

1- Why is $\mathbb{Z}G$ a free module over $\mathbb{Z}H$?

2- Why does every free $\mathbb{Z}G$-module is also free $\mathbb{Z}H$-module?

Answer 1

Note that you have $G = \bigcup_{g \in S} Hg$ with $S$ being a set of representatives for the right cosets of $G$ with respect to $H$. Now, we have $$\mathbb{Z} G = \bigoplus_{g \in S} \mathbb{Z} H g$$ and this is a decomposition of $\mathbb{Z} H$-modules. Note that $\mathbb{Z} H g \cong \mathbb{Z} H$ as $\mathbb{Z} H$-modules via multiplication with $g^{-1}$ from the right. It follows that $\mathbb{Z}G$ is the direct sum of copies of $\mathbb{Z}H$ and thus $\mathbb{Z}G$ is free as a $\mathbb{Z}H$-module.

For the second part note that every free $\mathbb{Z}G$-module is a direct sum of copis of $\mathbb{Z}G$ so the same argument applies.