Rational group algebras and maximal orders

Let $G$ be a finite group, and $\mathbb{Q}[G]$ be the rational group algebra. Then the group ring $\mathbb{Z}[G]$ is an order in $\mathbb{Q}[G]$, but is not in general a maximal order.

What can we say about the maximal order, if I've been given a specific $G$ and want to write down the maximal order of $\mathbb{Q}[G]$?

I am especially interested in abelian $G$, in particular $G$ cyclic, so I'd be maximally happy with a specific answer for $G$ cyclic.

Some other questions: Does every maximal order contain all the central idempotents of $\mathbb{Q}[G]$? Do such central idempotents then usually generate the maximal order over $\mathbb{Z}[G]$?

If $G$ is cyclic of order $n$, then $\Bbb{Q}[G]\cong \Bbb{Q}[x]/\langle x^n-1\rangle$. The polynomial $x^n-1=\prod_{d\mid n}\Phi_n(x)$ is a product of pairwise distinct irreducible cyclotomic polynomials $\Phi_n(x)$. By the Chinese remainder theorem we thus have $$\Bbb{Q}[G]\cong\bigoplus_{d\mid n}\Bbb{Q}[x]/\langle \Phi_n[x]\rangle\cong\bigoplus_{d\mid n}\Bbb{Q}[\zeta_d],$$ and the cyclotomic fields $\Bbb{Q}[\zeta_d]$ are the Wedderburn components of the group algebra. IIRC all maximal orders are gotten as direct sums of maximal orders of the Wedderburn components (please check this). In this case they are unique and equal the rings of integers of the component fields.
Whenever $G$ is abelian the Wedderburn components are still all fields and the same holds. We get a maximal order as a direct sum of rings of integers of the component fields.
In general we get that $$\Bbb{Q}[G]\cong\bigoplus_iM_{n_i}(D_i),$$ where $D_i$ are some division algebras with centers $F_i$ are that are finite extensions of $\Bbb{Q}$. Still (check Reiner again) we get maximal orders by putting together orders of the form $M_{n_i}(\Gamma_i)$, where $\Gamma_i$ is a maximal order of $D_i$. However, in the non-commutative case $\Gamma_i$ is not unique.