Centralizer of a subset ($C_G(A)={\bigcap }_{a\in A}{C}_G(a)$)

The centralizer of a subset $A$ of a group consists of the elements commuting with every member of $A$. It is the intersection of the element centralizers, hence a subgroup. Directly, identity commutes with every element, products of two elements that each commute with every member of $A$ still have that property, and the same is true of inverses. For $A=G$ this is the centre of a group. For an empty subset the commutation condition is vacuous and the centralizer is all of $G$.

Let a Hermitian matrix have distinct real eigenvalues with multiplicities $n_1,\dots ,n_k$. A unitary matrix commuting with it must preserve each eigenspace: if $Av=\lambda v$, then $A(Uv)=UAv=\lambda Uv$. Conversely, any unitary transformation acting independently on these mutually orthogonal eigenspaces commutes with $A$. The centralizer in $U(n)$ is therefore ${\prod }_{j}U(n_j)$. Imposing total determinant one gives the displayed subgroup inside the special unitary group $SU(n)$.

For multiplicities $(2,1)$ in SU(3), it is $S(U(2)\times U(1))\cong U(2)$, embedded by $U\mapsto \mathrm{diag}(U,(\det U{)}^{-1})$. Three distinct eigenvalues leave the diagonal maximal torus $U(1{)}^2$. This determines the unbroken group for a diagonal adjoint scalar field vacuum, including its global determinant constraint.