Show that SL(2,R), the multiplicative group of 2×2 matrices over the real numbers of determinant 1, is a normal subgroup of

Excerpt
Let K be its kernel, the set of all elements of G such that f(x)=e. To do ts, let a and b be in K and let g be any element of G. Being in K means f(a)=f(b)=e. Make calculations to prove f(ab)=e f(a^{-1})=e f(gag^{-1})=e.

Show that SL(2,R), the multiplicative group of 2×2 matrices over the real numbers of determinant 1, is a normal subgroup of GL(2,R), the multiplicative group of all 2×2 matrices over the real number wch are nonsingular. The simplest way to do ts is to give a group homomorpsm such that the former group is the kernel. Note that the determinant has the property det(AB)=det(A)det(B). 2.Tell whether the multiplicative group of nonsingular upper triangular matrices, that is, matrices like a b 0 d where the determinant ad is nonzero, is a normal subgroup of the group GL(2,R) of all nonsingular 2×2 matrices over the real numbers under the operation of multiplication. One way to do ts is to take the particular matrix 0 1 1 0 and conjugate the matrix in the previous display by ts matrix and see if the form is preserved. 3.If H and K are normal subgroups of G, show that their intersection is also a normal subgroup. To do ts, let b be an element of the intersection, so b is in H and b is in K. Then what can we say about gbg^{-1} in because b is in the normal subgroup K? What can we say because b is in the normal subgroup H? Why then is gbg^{-1} in the intersection of H and K? 4.Let G be the cyclic group Z_4 under addition, of integers modulo 4. Let H be the subgroup of multiples of 2, {0,2}. Write out the two distinct cosets of H. One is 0+H=H. The other is some g+H. Then write out the addition table for the quotient group G/H consisting of those two cosets. The sum of a+H and b+H will be a+b+H. 5.Let f:G to H be any group homomorpsm, that is, f(x)f(y) is always f(xy). It will follow that also f(x^{-1})=(f(x))^{-1}. Let K be its kernel, the set of all elements of G such that f(x)=e. To do ts, let a and b be in K and let g be any element of G. Being in K means f(a)=f(b)=e. Make calculations to prove f(ab)=e f(a^{-1})=e f(gag^{-1})=e.

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