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

This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Special linear group (?)) satisfying a particular subgroup property (namely, Cocentral subgroup (?)) in a particular group or type of group (namely, General linear group (?)).

## 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.

## Facts used

1. 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.