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General linear groups of dimension at least 2 are nonabelian


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.4 Exercise 1.4.8

Show that GLn(F) is nonabelian for all n2 and all fields F.


Solution: Recall that every field contains 0 and 1, and that 01. Suppose now that A,B are matrices in GLn(F) such that the first row of A is [1,0,,0,1], the first column of A is [1,0,,0,0], the first row of B is [1,0,,0,0], and the first column of B is [1,0,,0,1]. Such matrices always exist in GLn(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 ABBA. Thus GLn(F) is nonabelian for n2 and for all fields F where 01.


Linearity

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