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Consider these subgroups of $Gl_n(F)$, where $F$ is a field and $n \epsilon \Bbb N$

  • $Sl_n(F)$
  • $Diag_n(F)$
  • $F^* I_n$ ($F^*$ times $I_n$, i.e. all $f \epsilon F^*$ times $I_n$)


  1. Show that $Sl_n(F)$ is normal in $Gl_n(F)$
  2. Show that $F^* I_n$ is normal in $Gl_n(F)$
  3. When specifically is $Diag_n(F)$ normal in $Gl_n(F)$? (note: not true in general)
  4. When specifically is $Gl_n(F) \cong Sl_n(F) \times F^* I_n $? (also not true generally)

Comments: (1) easy. (4) My initial thoughts is true when n is odd, since when n is even in $\Bbb R$, $\Bbb R \bigcap I_n$ is multivalued.

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1. Take $S \epsilon Sl_n(F)$. Then for all $g \epsilon Gl_n(F)$ the $det(gS{g^-}^1) = det(g)det(S){det(g)^-}^1 = det(S) = 1$ therefore $gS{g^-}^1 \epsilon $ $Sl_n(F)$ $\Rightarrow$ $Sl_n(F)$ is normal. – KUSH Oct 31 '12 at 5:22
(4) Construct a projection $\operatorname{GL}_n F \to \operatorname{SL}_n F$, use this: – Alexei Averchenko Oct 31 '12 at 5:43
(1) Is $\operatorname{SL}_n$ a kernel of some group homomorphism? – Alexei Averchenko Oct 31 '12 at 5:49
(1, continued) As a side quest, prove that despite its ridiculously looking definition, this homomorphism is a natural transformation. – Alexei Averchenko Oct 31 '12 at 5:53
I have observed the map $$Sl_n(F) \mapsto Gl_n(F) \mapsto F^*$$ Where $Gl_n(F) \mapsto F^*$ through determinant map. – KUSH Oct 31 '12 at 6:04
up vote 1 down vote accepted

For (4), note that $GL_n(F)$ has all possible determinants whereas the right side has only $n$th powers as determinants. So a necessary condition for (4) is that every element of $F$ must have an $n$th root in $F$. Furthermore, in order for (4) to hold, $SL_n(F) \cap F^*I_n$ must consist only of the identity. So every element of $F$ must have a unique $n$th root in $F$. You can check that this necessary condition is also sufficient for (4).

Some examples of $(F,n)$ where this is satisfied: When $F=\mathbb{R}$ and $n$ is odd (which you sort of alluded to in your original post); when $F$ is a finite field of characteristic $p$, and $n$ is a power of $p$.

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Thanks, I hadn't considered finite fields like that, very helpful. – KUSH Oct 31 '12 at 6:07
I don't get it :( $\operatorname{GL}_n F$ retracts on $\operatorname{SL}_n F$ by dividing by the determinant, what prevents it from being a homomorphism in the cases that you presented? I mean, the short exact sequence has to not left-split, or else it's a direct product, right? – Alexei Averchenko Oct 31 '12 at 6:08
@Alexei It can't be a direct product if the intersection of $SL_n(F)$ and $F^*$ is nontrivial, for example. – Ted Oct 31 '12 at 7:22

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