I have to prove that there does not exist a surjective group algebra homomorphism from $FS_5$(the group algebra of the symmetric grpoup, $S_5$, over the field $F$) to $M_6(F)$, where $F$ is the field $\mathbb{Z}_2$ and $M_6(F)$ denotes the matrix algebra of $6\times 6$ matrices over the field $F$.

I have no idea how to prove it exactly. I am thinking which matrix doesn’t comes in range if particular map is defined. The dimension of domain algebra also bigger one. I already have link of the problem Artin-Wedderburn decomposition of $\mathbb{F}_2[S_5]/J$. But I do not know representation theory. Please give me a suggestion that does not use representation theory. Thanks.

  • $\begingroup$ Can you recall what $FS_5$ is? $\endgroup$ – mathcounterexamples.net Jan 1 at 5:22
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    $\begingroup$ I think the group algebra on the symmetric group $S_5$ over the field $F$. $\endgroup$ – mouthetics Jan 1 at 5:48
  • $\begingroup$ @mathcounterexamples.net Yes i think i already told about group algebra... $\endgroup$ – neelkanth Jan 1 at 5:53
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    $\begingroup$ If there was such a map, there would have to be an irreducible six-dimensional representation of $S_5$ over $F$. Can you determine the dimensions of the irreducible representations? $\endgroup$ – Lord Shark the Unknown Jan 1 at 6:37
  • $\begingroup$ @LordSharktheUnknown I am not having knowledge of Representation theory... $\endgroup$ – neelkanth Jan 1 at 6:38

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