Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.4 Exercise 1.4.8
Show that is nonabelian for all and all fields .
Solution: Recall that every field contains 0 and 1, and that . Suppose now that are matrices in such that the first row of is , the first column of is , the first row of is , and the first column of is . Such matrices always exist in ; for instance, take the identity matrix and change the or 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 and are invertible.
With and having this form, the -entry of is 2 and the -entry of is 1. If in then we have , a contradiction. Since matrices are equal precisely when their corresponding entries are equal, we have . Thus is nonabelian for and for all fields where .