Reduced Group algebras Take a finite group and a field of characteristic zero. The group algebra is due to Maschke's theorem semisimple so that its a finite direct sum of matrix algebras over division algebras. I like to know when the case occurs that the decomposition is a direct sum of division algebras (such algebras are called reduced which is equivalent to that there are no nilpotent elements). Clearly this is the case when the group is abelian. Is there any other example?
In the case of rational group algebras over dihedral and quanternion groups there is always a matrix algebra involved.
Of course in the case of complex numbers the answer is also no.
 A: The real group algebra of the quaternion group of order $8$ is the direct sum of $4$ copies of the real field and one copy of the $4$-dimensional division algebra of real quaternions.
Later edit : Here is a  proof of the claim that a finite group $G$ with a reduced group algebra $KG$ for $K$ a field of characteristic zero is Hamiltonian. By induction, we may suppose that $G/N$ is Hamiltonian whenever $1 \neq N \lhd G.$ 
Suppose first that $G$ has a unique minimal normal subgroup $M.$ Then $G$ has a  faithful irreducible $KG$-module $V$, and $G$ embeds in the multiplicative group of a division algebra. Now every Sylow subgroup of $G$ is cyclic or generalized quaternion, and if $G$ has even order, then $G$ has a unique involution. If $G$ has odd order, then $G$ has a normal Sylow $p$-subgroup for some prime $p.$ In all cases, $M$ must be cyclic of prime order, $q$ say. Now $C_{G}(M)$ is nilpotent, as $G/M$ is nilpotent. Let $r$ be a prime divisor of $[G:C_{G}(M)].$ Then $r$ divides $q-1$, and all subgroups of $G$ of order $qr$ are Abelian by a result of Burnside. Hence $C_{G}(M)$ contains a cyclic subgroup of order $qr$ containing $M.$ Since $G/M$ is Hamiltonian, this subgroup is normal in $G,$ so $G$ has a normal subgroup of order $r$. This contradicts the fact that $M$ is the unique minimal normal subgroup of $G.$ Hence $M \leq Z(G),$ and $G$ is nilpotent.
If $G$ has two minimal normal subgroups $L$ and $T,$ then $G/L$ and $G/T$ are both nilpotent, and hence so is $G.$   
In any case, then, $G$ is nilpotent, and all minimal normal subgroups of $G$ are central, and of prime order. Let $H$ be a subgroup of $G$ which is not normal. Then $H$ is corefree, for if $C = \cap_{g \in G}H^{g},$ and $C \neq 1,$ then $G/C$ is Hamiltonian and $H \lhd G.$
Let $x \in H$ be of prime order. Then whenever $1 \neq N \lhd G,$ we have that $xN \in Z(G/N)$. Hence $[G, \langle x \rangle ] \leq N.$ Hence $G$ must indeed have a unique minimal normal subgroup, for otherwise $x \in Z(G),$ a contradiction. Since $G$ is nilpotent, $G$ is now a $q$-group for some prime $q.$ Also, $G$ embeds in the multiplicative group of a division algebra, so $G$ is cyclic or generalized quaternion.
Hence we may suppose that $q =2,$ and that $G$ is generalized quaternion. If $|G| >8,$ then $G$ has a dihedral group of order $8$ as a homomorphic image, a contradiction, as a dihedral group of order $8$ is not Hamiltonian. Hence $G$ is either cyclic or quaternion of order $8,$ and the proof is complete.
