Wedderburn Decomposition by using Clifford theorem.

Question

Let $H$ be a normal subgroup of $G$ and we know Wedderburn decomposition of semi simple algebra $FH$ over a finite field $F$ as $$FH=F\oplus M_{3}(F)\oplus M_{3}(F)\oplus M_{4}(F)\oplus M_{5}(F).$$ But i want Wedderburn decomposition of $FG$. After my all calculation i found that $$FG=F\oplus M_{n_1}(F)\oplus M_{n_2}(F)\oplus M_{n_3}(F)\oplus M_{n_4}(F)$$ and two choices of $n_{i}$ as $2,3,4,5$ and $4,4,5,5,,6.$ Now i am confused which one is answer. I thought Clifford theorem will help but i am unable to apply it. Please help me apply to it or any other idea. Thanks.