Editing Unitary group (section)

== Properties ==
Since the [[determinant]] of a unitary matrix is a complex number with norm 1, the determinant gives a [[group homomorphism]]
: <math>\det \colon \operatorname{U}(n) \to \operatorname{U}(1).</math>

The [[kernel (group theory)|kernel]] of this homomorphism is the set of unitary matrices with determinant 1. This subgroup is called the '''[[special unitary group]]''', denoted SU(''n''). We then have a [[short exact sequence]] of Lie groups:
: <math>1 \to \operatorname{SU}(n) \to \operatorname{U}(n) \to \operatorname{U}(1) \to 1.</math>

The above map U(''n'') to U(1) has a section: we can view U(1) as the subgroup of U(''n'') that are diagonal with ''e<sup>iθ</sup>'' in the upper left corner and 1 on the rest of the diagonal. Therefore U(''n'') is a [[semidirect product]] of U(1) with SU(''n'').

The unitary group U(''n'') is not [[abelian group|abelian]] for {{nowrap|''n'' > 1}}. The [[center of a group|center]] of U(''n'') is the set of scalar matrices ''λI'' with {{nowrap|''λ'' ∈ U(1)}}; this follows from [[Schur's lemma]]. The center is then isomorphic to U(1). Since the center of U(''n'') is a 1-dimensional abelian [[normal subgroup]] of U(''n''), the unitary group is not [[Semisimple algebraic group|semisimple]], but it is [[Reductive group|reductive]].