Show that $\dim H^1(\Gamma(11)) = 3$

Question

Let $\Gamma(11) \subset \text{SL}(\mathbb{Z})$ be the invertible $2 \times 2$ matrices with one of the entries $0 \pmod {11}$. A congruence group is defined by: $$ \Gamma(11) = \left\{ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right): \left|\begin{array}{cc} a & b \\ c & d \end{array} \right| = ad-bc = 1 \text{ and } 11 \big| c\right\} $$ The group cohomology can be thought of as the homomorphisms from $\Gamma(11) \to \mathbb{Z}$. In our case, $\dim H^1(\Gamma_{11}) = 3$ here is one element: $$ \phi: \left( \begin{array}{cc} 7 & 5 \\ 11 & 8 \end{array}\right) \mapsto 1 \text{ and } \left( \begin{array}{cc} 3 & 1 \\ 11 & 3 \end{array} \right) \mapsto -1 \text{ and }\left( \begin{array}{cc} 127 & 150 \\ 11 & 3 \end{array} \right) \mapsto 2$$ This might be the group $(\mathbb{Z}, +)$ just the additive group on integers. One obvious homomorphism from $SL_2(\mathbb{Z}) \to \mathbb{Z}^\times$ is $\det \mapsto \{ \pm 1\}$. I am looking for a working defintion of first group cohomology.