Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.4 Exercise 1.4.8
Show that $GL_n(F)$ is nonabelian for all $n \geq 2$ and all fields $F$.
Solution: Recall that every field contains 0 and 1, and that $0 \neq 1$. Suppose now that $A, B$ are matrices in $GL_n(F)$ such that the first row of $A$ is $[1,0,\ldots,0,1]$, the first column of $A$ is $[1,0,\ldots,0,0]$, the first row of $B$ is $[1,0,\ldots,0,0]$, and the first column of $B$ is $[1,0,\ldots,0,1]$. Such matrices always exist in $GL_n(F)$; for instance, take the identity matrix and change the $(1,n)$ or $(n,1)$ entry from 0 to 1. The resulting matrix is either upper or lower triangular, so that the determinant is the product of the diagonal entries. This product is 1, so that $A$ and $B$ are invertible.
With $A$ and $B$ having this form, the $(1,1)$-entry of $AB$ is 2 and the $(1,1)$-entry of $BA$ is 1. If $1 = 2$ in $F$ then we have $0 = 1$, a contradiction. Since matrices are equal precisely when their corresponding entries are equal, we have $AB \neq BA$. Thus $GL_n(F)$ is nonabelian for $n \geq 2$ and for all fields $F$ where $0 \neq 1$.