# Show that U(n) / Z(U(n)) = SU(n) / Z(SU(n))?

• RickilusSmith
SU(n)/Z(SU(n)) is isomorphic to U(n)/Z(U(n)).In summary, the conversation discusses using the first isomorphism theorem to show that U(n) / Z(U(n)) is isomorphic to SU(n) / Z(SU(n)), with a proposed solution involving finding a homomorphism with both centers as the kernel. The suggestion is to draw a diagram and use the first isomorphism theorem to prove the isomorphism.
RickilusSmith

## Homework Statement

Show that U(n) / Z(U(n)) = SU(n) / Z(SU(n))?

## Homework Equations

Perhaps the first isomorphism theorem? That is, a group homomorphism ϕ of G onto G gives a 1-to-1 correspondence between G and G/(ker ϕ) that preserves products, that is, G is isomorphic to G/(ker ϕ)

## The Attempt at a Solution

I've been able to deduce the center of both U(n) and SU(n).

Z(U(n)) = { e^(i \theta) 1 : \theta \in R}

and

Z(SU(n)) = { e^(i \theta) 1 : e^(n i \theta) = 1}

From here, I've been trying to find a homomorphism with both centers as the kernel to another group. However, I've had no luck. I feel that after I'm able to find the group representation, proving that they are isomorphic, i.e. bijective and injective will be easy.

Thanks!

Perhaps you can draw as a nice diagram:

$$SU(n)/Z(SU(n))\leftarrow SU(n)\rightarrow U(n)\rightarrow U(n)/Z(U(n))$$

and check the properties of the arrows, kernels etc. See if you can add a unique arrow to make it commutative

## 1. What is U(n)?

U(n) is the group of all invertible n x n complex matrices under matrix multiplication.

## 2. What is Z(U(n))?

Z(U(n)), also known as the center of U(n), is the set of all matrices in U(n) that commute with every other matrix in U(n). In other words, it is the set of matrices that can be multiplied by any matrix in U(n) without changing the result.

## 3. What is SU(n)?

SU(n) is the special unitary group, which is a subgroup of U(n) consisting of matrices with unit determinant and complex entries.

## 4. What is Z(SU(n))?

Z(SU(n)), also known as the center of SU(n), is the intersection of SU(n) with Z(U(n)). In other words, it is the set of matrices with unit determinant that commute with every other matrix in SU(n).

## 5. How do you show that U(n) / Z(U(n)) = SU(n) / Z(SU(n))?

To show this, we must show that the quotient groups U(n) / Z(U(n)) and SU(n) / Z(SU(n)) are isomorphic. This can be done by constructing an isomorphism between the two groups, which is a bijective homomorphism. Specifically, we can define the map f: U(n) / Z(U(n)) → SU(n) / Z(SU(n)), where f([A]) = [det(A)*A], where [A] represents the coset of A in the quotient group. This map is well-defined, since multiplying a matrix by a scalar does not change its coset in the quotient group. It is also bijective and a homomorphism, which can be shown using the properties of determinants. Therefore, U(n) / Z(U(n)) and SU(n) / Z(SU(n)) are isomorphic, and thus equal.

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