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In a lot of references, people mention that $\operatorname{SL}_n(k)$ is a normal subgroup of $\operatorname{GL}_n(k)$ and $\operatorname{GL}_n(k)/\operatorname{SL}_n(k) \cong k^\times$ via the determinant map.

But I wonder if there is anything wrong with the following similar map

$$\operatorname{GL}_n(k)\to\operatorname{SL}_n(k), A\mapsto \frac{1}{\det A}A$$

This map is well-defined since $\det A\neq 0$ when $A\in\operatorname{GL}_n(k)$. The multiplication is preserved since $\det AB=\det A \det B$. This map is clearly surjective. So it induces $\operatorname{GL}_n(k)/k^\times \cong\operatorname{SL}_n(k)$.

Is there anything wrong here? I'm just surprised that I cannot find any source mentionging this one.

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  • $\begingroup$ $GL_n(k)/k^\times$ doesn't quite make sense. One can only mod out by a normal subgroup. $\endgroup$
    – ndhanson3
    Commented Feb 20, 2021 at 5:08
  • $\begingroup$ @ndhanson3 You can just identify $k^\times$ with the nonzero scalar matrices: $a \mapsto aI$. $\endgroup$
    – Viktor Vaughn
    Commented Feb 20, 2021 at 5:10
  • $\begingroup$ yeah and that one commutes with everything so that's definitely a normal subgroup $\endgroup$
    – T C
    Commented Feb 20, 2021 at 5:11
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    $\begingroup$ $(\det A)^{-1}A$ need not have determinant 1, since $\det(aA)=a^n\det A$. $\endgroup$
    – user10354138
    Commented Feb 20, 2021 at 5:17
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    $\begingroup$ @ndhanson3 No. Assuming you mean $(\det A)^{-1/n}A$ you have problem with (1) there may not be an $n$-th root of $\det A$; and (2) arbitrarily choosing $n$-th roots would spoil the homomorphism property. $\endgroup$
    – user10354138
    Commented Feb 20, 2021 at 5:21

1 Answer 1

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As pointed out in the comments, the given map does not map into $SL_n(k)$, and in general these groups are not isomorphic. For instance, if we consider the case of Lie groups ($k=\mathbb{C}$ or $k=\mathbb{R}$) then there is at most one map with prescribed differential at the identity. Any isomorphism would have the identity as differential at the identity, like the natural map $SL_n(\mathbb{C}) \to GL_n(\mathbb{C})/k^*$. However, this map has a finite kernel, namely the $n$th roots of unity times the identity (if $k=\mathbb{R}$, take $n$ even and $\pm I$). By uniqueness of maps corresponding to identity differential, there is no isomorphism between the groups.

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