Question

Let R be any ring with identity 1 and GL (n,R) the group of invertible n x n matrices over R and SL (n, R) those matrices with determinant 1. Show that SL (n, R) forms a normal subgroup of GL (n,R)

Answer 1

First of all, its a subgroup because the determinant of a product is the product of the determinants, and determinant of the inverse is 1/determinant.

Why is it normal? Conjugate an arbitrary element of SL(n,R) by an element of GL(n,R) and show that it's still in SL(n,R). This follows the two properties of the determinant I just cited. Let me know if you're still stuck.

Answer 2

Great hint commutant! remember also that \(\text{det}(AB)=\det(A)\det(B)\).

Answer 3

Also? I hope that property was clear from what I said.

Answer 4

sorry commutant. It seems I didn't read your answer carefully :D.