0
$\begingroup$

Could someone help with a proof: $F^*$ $I_n$ is normal in $GL_n(F)$.


Notice GL maps to F*:

$det:GL_n(F) \mapsto F^*$ by the determinant map.

And notice it maps back:

$F^* \mapsto GL_n(F)$ by ($a \mapsto aI_n$) where $a \epsilon F^*$

Also note:

Not dot product (so $F^*I_n$ is not a scalar vector, it is a diagonal matrix)

1 Answers 1

1

If by $\,\Bbb F^*\cdot I_n\,$ you mean the subgroup of all the non-zero scalar matrices then just check that this subgroup is just the center of $\,GL_n(\Bbb F)\,$...

  • 0
    Actually no sorry not $F^*$ dot $I_n$. It would be The matrix with only diagonal entries, like for instance if n=3 then: $[a 0 0] [0 a 0] [0 0 a]$ where $a \epsilon F*$2012-11-01
  • 1
    Well, *actually* your example **is** a scalar matrix...2012-11-01
  • 1
    And nobody meant that $\,\Bbb F^*\cdot I_n\,$ is "dot product".2012-11-01