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# Show that SL(2,R) ◅ GL(2,R)?

SL(2,R) is within GL(2,R) for all ad-bc = 1

### 1 Answer

- YoranLv 58 years agoFavorite Answer
GL(n,R) is the group of nxn invertible matrices under matrix multiplication, and SL(n,R) is the group of nxn invertible matrices with determinant 1.

Now first we have to show that SL(2,R) is a subgroup of GL(2,R). Clearly the identity element is in SL(2,R) : ((1,0) (0,1)) has determinant 1. If we have two elements A and B in SL(2,R), then their product is again an invertible nxn matrix, which has determinant:

det(AB) = det(A)det(B) = 1. So it's in SL(2,R). The inverse of an A in SL(2,R) is again an nxn invertible matrix A^-1 with determinant det(A^-1) = det(A)^-1 = 1^-1 =1.

So SL(2,R) is a subgroup of GL(2,R).

In order to check that it's normal in GL(2,R), note that if S is in SL(2,R) and G in GL(2,R) :

GSG^-1 is again an invertible nxn matrix, with determinant:

det(GSG^-1) = det(G)det(S)det(G)^-1 = det(G)det(G)^-1det(S) = det(S) = 1.

So it's again in SL(2,R).

This shows that SL(2,R) is a normal subgroup of GL(2,R).