anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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.
watchmath
  • watchmath
Great hint commutant! remember also that \(\text{det}(AB)=\det(A)\det(B)\).
anonymous
  • anonymous
Also? I hope that property was clear from what I said.

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

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