Special linear group is cocentral in general linear group iff nth power map is surjective

Statement
Let $$k$$ be a field and $$n$$ be a natural number. Let $$GL_n(k)$$ be the group of invertible $$n \times n$$ mtarices over $$k$$, and $$SL_n(k)$$ be the special linear group: the subgroup comprising matrices of determinant one.

If the map $$x \mapsto x^n$$ is a surjective map from $$k$$ to $$k$$, then $$SL_n(k)$$ is a cocentral subgroup of $$GL_n(k)$$: its product with the center of $$GL_n(k)$$ equals $$GL_n(k)$$. In particular:


 * For $$k$$ a finite field of order $$q$$, the $$n^{th}$$ power map is surjective if and only if $$n$$ is relatively prime to $$q - 1$$.
 * For $$k = \R$$, the $$n^{th}$$ power map is surjective if and only if $$n$$ is odd.
 * For $$k$$ an algebraically closed field, the $$n^{th}$$ power map is always surjective.

Corollaries

 * Projective special linear group equals projective general linear group iff nth power map is surjective

Other related facts

 * Center of general linear group equals group of scalar matrices over center

Facts used

 * 1) uses::center of general linear group is group of scalar matrices over center

Proof
Since $$SL_n(k)$$ is the kernel of the determinant homomorphism, it suffices to show that the image of the center of $$GL_n(k)$$, under the determinant homomorphism, equals the whole of $$k^*$$.

By fact (1), the center of $$GL_n(k)$$ is the group of scalar matrices with scalar entry a nonzero element of $$k$$. If this nonzero element is $$\lambda$$, the determinant is $$\lambda^n$$. Thus, the map to $$k^*$$ is surjective if and only if every nonzero element is the $$n^{th}$$ power of some nonzero element, which happens if and only if the $$n^{th}$$ power map is surjective.