**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.4 Exercise 1.4.5**

Let $F$ be a field. Show that $GL_n(F)$ is a finite group if and only if $F$ is finite.

Solution: Suppose $F$ is finite. Then there are only finitely many $n \times n$ matrices over $F$, in particular, $|F|^{n^2}$. Thus there are at most $|F|^{n^2}$ elements in $GL_n(F)$.

Suppose now that $F$ is infinite. Note that for all $\alpha \in F$, the matrix $A_\alpha$ defined such that $a_{i,j} = \alpha$ if $i = j = 1$, 1 if $i = j \neq 1$, and $0$ otherwise has determinant $\alpha$ and so is in $GL_n(F)$. Thus $GL_n(F)$ is infinite.