Irreducible representation of Hecke algebra are dimension 1 or 2 First, Denote by $H(2)$ the algebra generated by the elements $Y_{1}, Y_{2}$, and $s$ subject to the following relations:
$$
s^{2}=1, \quad Y_{1} Y_{2}=Y_{2} Y_{1}, \quad s Y_{1}+1=Y_{2} s
$$
My goal is to show that finite irreducible representations of H(2) are either one dimensional or two dimensional.
Since $Y_{1}$ and $Y_{2}$ commute, we know that they have a common eigenbasis.
I would show that an irreducible representation of H(2) is generated by (v, sv) where v is an eigenvector of $Y_{1}$ and $Y_{2}$ but I don't know how to do it.Thanks.

It is not necessary that $Y_1$ and $Y_2$ have a common eigenbasis, because $Y_1$ and $Y_2$ might not be diagonalizable at all! However, you are right that this is a key observation, and what is true is that $Y_1$ and $Y_2$ will have a common eigenvector.
For simplicity, suppose that $V$ is an $H(2)$-module such that $s$ acts by the identity. Then $Y_2$ acts by $Y_1+1$. In this case, can you show that, if $v$ is an eigenvector of $Y_1$, then the line spanned by $v$ is an $H(2)$-submodule of $V$? What about if $s$ acts by $-1$?
Can you see how to generalize this to the case of $V$ where $s$ does not act by a scalar? Perhaps you can guess what a basis of a two-dimensional $H(2)$-submodule of $V$ could be in this case.
