Center of finitely generated $C^\ast$-algebra

Let $\mathcal{A}$ be a $C^\ast$-algebra finitely generated by $n$ elements $\{a_1,\dots,a_n\}$.

Elements in the center $Z(\mathcal{A})$ commute with each of the generators $$Z(\mathcal{A})\subset\{a\in\mathcal{A}:aa_i=a_ia\}$$ What about vice versa: $$\{a\in\mathcal{A}:aa_i=a_ia\}\subset Z(\mathcal{A})?$$

If $a\in \mathcal{A}$ is non-central does $aa_i\neq a_ia$ for some generator $a_i$?

It would seem that a proof by contradiction might do it.

Solution 1:

Let $a \in \mathcal A$ be an element commuting with all the generators. The centralizer of $a$ is defined as : $$Z_\mathcal A(a) = \{b\in A | ab = ba\}$$ It a sub-$C^*$-algebra of $\mathcal A$ (in fact it is the largest sub-$C^*$-algebra containing $a$ in it's center). Since $Z_\mathcal A(a)$ contains the generators of $\mathcal A$, it is equal to $\mathcal A$ and $a \in Z(\mathcal A)$.

Therefore : $$Z (\mathcal A) = \{ a\in \mathcal A|\forall i\in \{1,\ldots,n \}, a_i a= aa_i\}$$