For which finite groups $G$ does the Wedderburn decomposition of $\mathbb{Q}[G]$ consist only of fields and division algebras?

Question

Let $G$ be a finite group. Then the rational group algebra $\mathbb{Q}[G]$ has a wedderburn decomposition of the form $\prod_i M_{n_i}(D_i)$ where each $D_i$ is a division algebra.

My question is: for which $G$ do we have $n_i=1$ for all $i$? In other words, for which finite groups $G$ does the Wedderburn decomposition of $\mathbb{Q}[G]$ consist only of fields and division algebras?

Clearly, this is true when $G$ is abelian. Another example is $G=Q_8$, the quaternion group of order $8$. Are there other examples? Is there a complete classification of such groups? If so, can you provide a reference?

Answer 1

The groups you are looking for are precisely those for which the group algebra $\mathbb Q[G]$ does not contain nonzero nilpotent elements. These groups have been classified by Sehgal in here. The finite groups which have this property are the abelian ones and the Hamiltonian groups of order $2^mt$ (where $t$ is odd), such that $2$ has odd order modulo $t$.

Answer 2

The condition is clearly equivalent to require that the group algebra $\mathbb{Q}G$ has no nonzero nilpotent elements and there is already an answer here:

https://math.stackexchange.com/questions/906764/reduced-group-algebras

Answer 3

A classification of finite subgroups of multiplicative groups of division algebras over $\mathbb{Q}$ was given by Fein and Schacher (Sect 2. there), building on work by Amitsur. In particular such a group is either soluble, of a restricted set of types listed there, or $SL(2,5)$.

So the groups $S$ you are after are built off these groups. However, you can prune that list further, by only taking these that are of the type you are after. E.g. $SL(2,5)$ need not be considered, as its quotient $A_5$ is not of your type.