Wedderburn Decomposition by using Clifford theorem.

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.

• Something is missing. It sounds like you are given the decomposition of $F[H]$ implying, among other things, that $|H|=60$ and that $F[H]$ is semisimple. Then you ask as to divine the decomposition of $F[G]$. Without telling us anything at all about $G$, for example its cardinality. With this data it is impossible to say even the dimension of $F[G]$. Were you also given that $F[G]$ has exactly five simple components? Or what? – Jyrki Lahtonen Dec 28 '18 at 7:34
• Also, why did you tag this with clifford-algebras? Clifford algebras are associative algebras constructed starting from a quadratic form on a vector space. – Jyrki Lahtonen Dec 28 '18 at 7:37
• Or, were you given those two choices for the Wedderburn decomposition of $G$, and asked to determine, which one may come from a group $G$ having $H$ as a subgroup. That would make sense, but you should not leave us guessing what is given, and what's not. – Jyrki Lahtonen Dec 28 '18 at 7:41
• @JyrkiLahtonen yes sir decomposition of normal subgroup is given... – neelkanth Dec 28 '18 at 15:02
• Sir H is of index $2$ i.e.\ $G$ is of order $120.$ – neelkanth Dec 28 '18 at 15:04