Use the isomorphism theorem to determine the group GL2RSL2(R). Here GL2(R) is the group of 2×2 matrices with determinant not equal to 0, and SL2(R) is the group of 2×2 matrices with determinant 1. In the first part of the problem, I proved that SL2(R) is a normal subgroup of GL2(R). Now it wants me to use the isomorphism theorem. I tried using|GL2RSL2(R)|=|GL2(R)||SL2(R)|,but since both groups have infinite order, I don't think I can use this here.

Question

Use the isomorphism theorem to determine the group GL2RSL2(R). Here GL2(R) is the group of 2×2 matrices with determinant not equal to 0, and SL2(R) is the group of 2×2 matrices with determinant 1. In the first part of the problem, I proved that SL2(R) is a normal subgroup of GL2(R). Now it wants me to use the isomorphism theorem. I tried using|GL2RSL2(R)|=|GL2(R)||SL2(R)|,but since both groups have infinite order, I don&#039;t think I can use this here.

cinereod3am · Accepted Answer

Step 1Define f: f:GL2(R)→R⋅ such thatf(M)=det(M)⏟OntoLet M1,M2∈GL2(R⋅) thenf(M1M2)=det(M1M2)=det(M1)det(M2)=f(M1)(M2)Step 2Hence f is homomorphism for ker(f)={M:M∈GL2(R)  such that  det(M)=1}=SL2(R)From first theorem in isomorphismGL2RSL2(R)≈R⋅

Use the isomorphism theorem to determine the group

Answered question

Answer & Explanation

New Questions in Abstract algebra