SU(n) is a simple group?
Yes, SU(n) is a simple group for all positive integers n. A simple group is a group that does not have any nontrivial normal subgroups. A normal subgroup is a subgroup that is invariant under conjugation. In other words, if H is a normal subgroup of G, then for any g in G, gH = Hg.
The group SU(n) is the group of all n × n unitary matrices with determinant 1. Unitary matrices are matrices that preserve the inner product of vectors. The determinant of a unitary matrix is always 1, so SU(n) is a subgroup of the group U(n) of all n × n unitary matrices.
To show that SU(n) is simple, we need to show that it does not have any nontrivial normal subgroups. Suppose that H is a nontrivial normal subgroup of SU(n). Then H contains a matrix U that is not the identity matrix. Since U is in H, it is also in U(n). But U(n) is a simple group, so H must be equal to U(n). This is a contradiction, since H is a proper subgroup of U(n). Therefore, SU(n) does not have any nontrivial normal