Let $\displaystyle G$ be a group and let $\displaystyle z$ be in the center of $\displaystyle G$.

Why is $\displaystyle \rho(z)$ for every irreducible representation $\displaystyle \rho$ of $\displaystyle G$ a scalair?

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- Dec 1st 2010, 01:56 AMbram kierkelsIrreducible representation
Let $\displaystyle G$ be a group and let $\displaystyle z$ be in the center of $\displaystyle G$.

Why is $\displaystyle \rho(z)$ for every irreducible representation $\displaystyle \rho$ of $\displaystyle G$ a scalair? - Dec 1st 2010, 01:00 PMDrexel28
- Dec 2nd 2010, 02:51 AMbram kierkels
rho is just an arbitrary representation, thus $\displaystyle \rho:G\rightarrow GL(V)$ is a homomorphism on a (complex) vector space $\displaystyle V$, where GL(V) is the general lineair group.

I think it is easier then it looks:

Since $\displaystyle \rho(z):V\rightarrow V$, and $\displaystyle \rho(z)\rho(x)=\rho(zx)=\rho(xz)=\rho(x)\rho(z).$, it follows directly from this corollary of Schur's Lemma:

PlanetMath: Schur's lemma