Isomorphism between linear groups when degree power map is bijective

Statement for an arbitrary field
Suppose $$K$$ is a field and $$n$$ is a natural number such that the map $$x \mapsto x^n$$ is a bijection from $$K$$ to itself. Then, the following are true:


 * 1) The general linear group $$GL(n,K)$$ is an internal direct product of these two subgroups: the special linear group $$SL(n,K)$$ and the center $$Z(GL(n,K))$$, which is the group of scalar matrices and is isomorphic to the multiplicative group $$K^*$$.
 * 2) The composite of the inclusion of $$SL(n,K)$$ in $$GL(n,K)$$ and the quotient map from $$GL(n,K)$$ to the projective general linear group $$PGL(n,K)$$ is an isomorphism.
 * 3) In particular, from (2), we get that $$SL(n,K) \cong PGL(n,K) \cong PSL(n,K)$$.

Statement for a finite field
Suppose $$q$$ is a prime power and $$n$$ is a natural number such that $$\operatorname{gcd}(n,q - 1) = 1$$. Then the field $$\mathbb{F}_q$$ of size $$q$$ satisfies the condition that the map $$x \mapsto x^n$$ is a bijection, and the following are true:


 * 1) The general linear group $$GL(n,q)$$ is an internal direct product of these two subgroups: the special linear group $$SL(n,q)$$ and the center $$Z(GL(n,q))$$, which is the group of scalar matrices and is isomorphic to the multiplicative group $$\mathbb{F}_q^*$$, which is a cyclic group of order $$q - 1$$ (see multiplicative group of a finite field is cyclic).
 * 2) The composite of the inclusion of $$SL(n,q)$$ in $$GL(n,q)$$ and the quotient map from $$GL(n,q)$$ to the projective general linear group $$PGL(n,q)$$ is an isomorphism.
 * 3) In particular, from (2), we get that $$SL(n,q) \cong PGL(n,q) \cong PSL(n,q)$$.

Related facts

 * Isomorphism between linear groups over field:F2