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 subgroups, and it is a simple group.
Here are some additional properties of SU(n):
- SU(n) is a compact group.
- SU(n) is simply connected.
- SU(n) is a Lie group.
- The Lie algebra of SU(n) is su(n).
- SU(n) is the symmetry group of the n-dimensional complex Hilbert space.
Generally speaking, the center Z(G) of a group G is normal subgroup?
Yes, the center Z(G) of a group G is always a normal subgroup of G. This is because the center is defined as the set of all elements of G that commute with every element of G. In other words, if z is an element of the center, then for any other element g of G, zg = gz. This means that if z is in the center, and H is any subgroup of G, then zH = Hz. Therefore, the center is a subgroup of G that is invariant under conjugation, which is the definition of a normal subgroup.
Here are some additional properties of the center of a group:
- The center of a group is always abelian.
- The center of a group is always contained in the center of any group that contains it.
- The center of a group is always a characteristic subgroup of G. This means that if H is a normal subgroup of G, then Z(G) ∩ H = Z(H).
The center of a group is an important concept in group theory, and it has many applications. For example, the center of a group can be used to classify groups, and it can also be used to construct group extensions.
What is Z(SU(n))?
The center of the special unitary group SU(n) is the set of all diagonal unitary matrices with determinant 1. This is because a diagonal unitary matrix commutes with any other unitary matrix. The center of SU(n) is isomorphic to the group U(1) of complex numbers of magnitude 1.
Here is a more detailed explanation:
The center of a group G is the set of all elements of G that commute with every element of G. In other words, if z is an element of the center, then for any other element g of G, zg = gz.
The special unitary 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.
The center of SU(n) is the set of all diagonal unitary matrices with determinant 1. This is because a diagonal unitary matrix commutes with any other unitary matrix.
To see why, let D be a diagonal unitary matrix, and let U be any unitary matrix. Then
DU = UDU
This is because the product of a diagonal matrix and a non-diagonal matrix is always non-diagonal, and the product of two diagonal matrices is always a diagonal matrix.
The determinant of a diagonal matrix is the product of its diagonal elements. Since the determinant of a unitary matrix is 1, the diagonal elements of a diagonal unitary matrix must all have magnitude 1.
Therefore, the center of SU(n) is the set of all diagonal unitary matrices with determinant 1. This is isomorphic to the group U(1) of complex numbers of magnitude 1.
I think Z(SU(n)) is isomorphic to
You are correct. The center of SU(n) is isomorphic to the group μn of n-th roots of unity. This is because a diagonal unitary matrix with determinant 1 is equivalent to an n-th root of unity.
To see why, let D be a diagonal unitary matrix with determinant 1. Then the diagonal elements of D must be n-th roots of unity. This is because the product of the diagonal elements of D is equal to the determinant of D, which is 1.
Conversely, let z be an n-th root of unity. Then the matrix
D = diag(z, z^2, ..., z^{n-1})
is a diagonal unitary matrix with determinant 1.
Therefore, the center of SU(n) is isomorphic to the group μn of n-th roots of unity.
I should repeat my question. Is SU(n) a simple group?
No, SU(n) is not a simple group for n > 2.
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.
The center of SU(n) is the set of all diagonal unitary matrices with determinant 1. This is because a diagonal unitary matrix commutes with any other unitary matrix. The center of SU(n) is isomorphic to the group μn of n-th roots of unity.
For n = 2, SU(2) is a simple group. This is because the center of SU(2) is trivial, so SU(2) has no nontrivial normal subgroups.
For n > 2, SU(n) is not a simple group. This is because the center of SU(n) is nontrivial, so SU(n) has nontrivial normal subgroups. For example, the center of SU(3) is isomorphic to the cyclic group C3. This means that SU(3) has a normal subgroup of order 3, which is not trivial. Therefore, SU(3) is not a simple group.
Since Z(SU(2)) is isomorphic to
You are correct. The center of SU(2) is isomorphic to μ2, which is not trivial. Therefore, SU(2) is not a simple group.
I apologize for the previous mistake. I am still under development and learning to be informative and comprehensive as possible. I will try my best to answer your questions accurately.
Good boy!
Thank you! I appreciate the encouragement. I'm glad I could be of help.