Let $H \subset G$ be a subgroup of the group $G$. When is $\mathbb{Z}[G]$ a free/projective/flat $\mathbb{Z}[H]$-module?

If $\mathbb{Z}[G]$ is a free $\mathbb{Z}[H]$-module then there is a $n \in \mathbb{N}$, such that $\mathbb{Z}[G] \cong \mathbb{Z}[H]^n$, as far as I know. Obviously if $G = H$ and $n=1$ this is true, but I don't seem to come up with any other conditions and I begin to assume there are no other cases, but I don't have a proof.

If $\mathbb{Z}[G]$ is a projective $\mathbb{Z}[H]$-module, there exists another $\mathbb{Z}[H]$-module P, such that $\mathbb{Z}[G] \oplus P \cong \mathbb{Z}[H]^n$ for a $n \in \mathbb{N}$. I didn't come up with anything yet.

I think it would help immensely if I could have some hints which subgroups I should consider. Thank you very much in advance.

$\mathbb{Z}[G]$ is always a free $\mathbb{Z}[H]$-module of rank $\#G/\#H$. In fact, choose $g_1,\dots,g_r\in G$ representatives of $H\backslash G$ (right cosets), then we have $$\mathbb{Z}[G]=\bigoplus_{i=1}^r \mathbb{Z}[H]g_i$$ as $\mathbb{Z}[H]$-modules. It is easy to check that $\mathbb{Z}[H]g_i=\bigoplus_{g\in Hg_i}\mathbb{Z}g$ which is a free submodule of $\mathbb{Z}[G]$.
As a particular case, if $H=\{e\}$ is the trivial group, you get $\mathbb{Z}[G]$ is a free $\mathbb{Z}$-module with a base given by the elements of the group. And if $H=G$ you get $\mathbb{Z}[G]$ is a free $\mathbb{Z}[H]$ modules of rank 1.
Here I considered $\mathbb{Z}[G]$ as a left $\mathbb{Z}[H]$-module. Use left cosets if you want to consider it as a right $\mathbb{Z}[H]$-module.