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The unitary irreducible representations of the group ${\rm SU(2)}$ are $(2j+1)$ dimensional where $j=0,1/2,1,2,3/2...$etc. Consider such a representation of dimension $3$ which corresponds to $j=1$. These representation matrices are three-dimensional and unitary. But do they also have determinant $+1$? If yes, why?

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Yes. Here is one argument:

Let $H=\ker(\det \circ \rho)$, where $\rho$ is your favorite (nontrivial) representation. The question is to show that $H={\bf{SU}}(2)$. Then $H$ is a normal subgroup, and ${\bf{SU}}(2)/H$ is an abelian group. There are very few normal subgroups (see Normal Subgroups of $SU(n)$), namely there are finite or ${\bf{SU}}(2)$ itself, but for a normal finite subgroup $H$ the quotient is never abelian.

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