# Show that U(n) / Z(U(n)) = SU(n) / Z(SU(n))? (1 Viewer)

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#### RickilusSmith

1. The problem statement, all variables and given/known data

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

2. Relevant 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)

3. 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

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