## General linear groups of dimension at least 2 are nonabelian

Show that is nonabelian for all and all fields .

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 A 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 BA 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 .

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## Comments

hola espero que esten bien

solo queria decirles que esta pagina es muy interesante los felicito.

por otro lado quisiera que me respondieran una pregunta referente al ejercicio 1.4 # 8 queria saber como asegurar que esas entradas que me dan para la matriz me forme una matriz invertible o no. gracias

Gracias por leer!

Mi español es muy malo, así que utilicé Google Translate. Si he entendido bien, no es inmediatamente obvio que las matrices invertibles A y B existen.

Sí, esto podría ser más claro. Las matrices A y B existe siempre, consideras la matriz identidad con dimensiones (n,n) que el (1, n) o (n, 1) entrada 1 en lugar de 0. Estas matrices son triangulares superior e inferior, respectivamente, por lo que su factor determinante es el producto de las entradas diagonales. Dado que el determinante es 1, A y B son invertibles.

He editado la solución para hacer de este explícito.

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Thank you for reading!

My spanish is very bad, so I used Google Translate. If I understand correctly, it is not immediately obvious that invertible matrices A and B exist.

Yes, this could be made more clear. Matrices A and B always exist; take the n by n identity matrix and make the (1,n) or (n,1) entry 1 instead of 0. These matrices are upper and lower triangular, respectively, so their determinant is the product of the diagonal entries. Since the determinant is 1, A and B are invertible.

I edited the solution to make this explicit.