If $x\in Z(\mathbb{C}[G])$, why is $x$ invertible in $Z(\mathbb{C}[G]e)$ if $xe=x$? I'm confused by something I'm reading. Suppose $G$ is a finite group, and you have a central element $x\in Z(\mathbb{C}[G])$. The center of the group algebra has a basis of central primitive idempotents indexed by the irreducible characters of $G$, call them $\{e_\chi\}_\chi$.
Suppose $e=\sum_\chi e_\chi$ where the sum is over the $\chi$ such that $xe_\chi\neq 0$. Then I believe it is the case that $x=xe$, since each $xe_\chi$ is the projection of $\chi$ onto the corresponding isotypic component, and by summing all of them we make sure to keep all of them. So $x\in Z(\mathbb{C}[G]e)$. But apparently $x\in Z(\mathbb{C}[G]e)^\times$, but I don't get why $x$ is also a unit here?
 A: This is easier to understand if you identify $\mathbb{C}[G]$ with a product $\prod_\chi M_{\chi(1)}(\mathbb{C})$ of matrix rings (one for each irreducible representation).  The center is then $\prod Z(M_{\chi(1)}(\mathbb{C}))$, but the center of a matrix ring just consists of the matrices of the form $cI$ for $c\in\mathbb{C}$.  So if $x\in\prod Z(M_\chi(1)(\mathbb{C}))$ then it is a scalar multiple of the identity matrix on each coordinate.  The ring $\mathbb{C}[G]e$ is then the product $\prod_{xe_\chi \neq 0}M_{\chi(1)}(\mathbb{C})$ of just those factors on which $x$ is nonzero.  So on each coordinate of $\mathbb{C}[G]e$, $x$ is a matrix of the form $cI$ where $c$ is a nonzero scalar.  Any such matrix is invertible (with inverse $c^{-1}I$), so $x$ is a unit in this ring.
To put it another way, by looking at one irreducible representation at a time, we can assume our ring is a matrix ring.  Then the statement just becomes that any nonzero element of the center of a matrix ring $M_n(\mathbb{C})$ is invertible, which is true since it must be $cI$ for some nonzero $c\in\mathbb{C}$.
