Show that SL(2,R), the multiplicative group of 2x2 matrices over the real numbers of determinant 1, is a normal subgroup of GL(2,R), the multiplicative group of all 2x2 matrices over the real number which are nonsingular.
- Show that SL(2,R), the multiplicative group of 2x2 matrices over the real numbers of determinant 1, is a normal subgroup of GL(2,R), the multiplicative group of all 2x2 matrices over the real number which are nonsingular.
The simplest way to do this is to give a group homomorphism 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 2x2 matrices over the real numbers under the operation of multiplication. One way to do this is to take the particular matrix
0 1
1 0
and conjugate the matrix in the previous display by this 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 this, 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 homomorphism, 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 this, 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.
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
