Character of $S_3$ I am trying to learn about the characters of a group but I think I am missing something.
Consider $S_3$. This has three elements which fix one thing, two elements which fix nothing and one element which fixes everything. So its character should be $\chi=(1,1,1,0,0,3)$ since the trace is just equal to the number of fixed elements (using the standard representation of a permutation matrix).
Now I think this is an irreducible representation, so $\langle\chi,\chi\rangle$ should be 1. But it's $\frac{1}{6}(1+1+1+9)=2$. 
So is the permutation matrix representation actually reducible? Or am I misunderstanding something?

For any $S_n$, the representation $\pi$ on ${\bf C}^n$ by permuting basis vectors is a product of the trivial representation $\pi\vert\{(x,x,\ldots,x)\vert x\in{\bf C}\}$ and the faithful representation $\pi\vert V$ with $V=\{(x_1,\ldots,x_n)\vert \sum x_j=0\}$, the latter being irreducible.
