Let $a,b,c$ be integers and let $[a,b,c]$ denote the mapping \begin{equation*} [a,b,c]\colon \Z \times \Z \to \Z, (x,y) \mapsto ax^2 +bxy +cy^2. \end{equation*}Let $G$ be the group of $2\times 2$ matrices such that $\det{A}=\pm 1\;\forall\;A\in G$ . The substitution \begin{equation*} \begin{vect} x \\ y \end{vect} \mapsto A \begin{vect} x \\ y \end{vect} \end{equation*}leads to \begin{equation*} [a,b,c](a_{11}x +a_{12}y, a_{21}x+a_{22}y) =a^{'}x^2+b^{'}xy+c^{'}y^2, \end{equation*}where \begin{eqnarray} \label{eq:1} a^{'}&=& aa_{11}^2 +ba_{11}a_{21} +ca_{21}^2 \\ \nonumber b^{'}&=& 2aa_{11}a_{12} +2ca_{21}a_{22} +b(a_{11}a_{22} +a_{12}a_{21}) \\ \nonumber c^{'}&=& aa_{12}^2 +ba_{12} a_{22} +ca_{22}^2 \end{eqnarray}So we define \begin{equation*} [a,b,c]\ast A :=[a^{'}, b^{'}, c^{'}] \end{equation*}to be the binary quadratic form with coefficients $a^{'},b^{'}, c^{'}$ of $x^2, xy, y^2$ , respectively as in (). Putting in $A=\begin{matrix} 1&0 \\ 0&1 \end{matrix}$ we have $[a,b,c]\ast A=[a,b,c]$ for any binary quadratic form $[a,b,c]$ . Now let $B$ be another matrix in $G$ . We must show that \begin{equation*} [a,b,c] \ast (AB)=([a,b,c] \ast A)\ast B. \end{equation*}Set $[a,b,c]\ast (AB):=[a^{''}, b^{''}, c^{''}]$ . So we have \begin{eqnarray} \label{eq:3} a^{''}&=&a\left(a_{11}b_{11} +a_{12}b_{21}\right)^2 +c\left(a_{21}b_{11} +a_{22}b_{21}\right)^2 +b\left(a_{11}b_{11} +a_{12}b_{21}\right)(a_{21}b_{11} +a_{22}b_{21}) \\ \nonumber &=&a^{'}b_{11}^2 +c^{'}b_{21}^2 +\left(2a\cdot a_{11}a_{12} +2c\cdot a_{21}a_{22}+b\left(a_{11}a_{22}+a_{12}a_{21}\right)\right)b_{11}b_{21} \\ c^{''}&=&a\left(a_{11}b_{12}+a_{12}b_{22}\right)^2+c\cdot \left(a_{21}b_{12}+a_{22}b_{22}\right)^2 +b\cdot \left(a_{11}b_{12}+a_{12}b_{22}\right)\left(a_{21}b_{12}+a_{22}b_{22}\right) \\ \nonumber &=&a^{'}b_{12}^2+c^{'}b_{22}^2+\left(2a\cdot a_{11}a_{12}+2ca_{21}a_{22} +b\left(a_{11}a_{22} +a_{12}a_{21}\right)\right)b_{12}b_{22} \end{eqnarray}as desired. For the coefficient $b^{''}$ we get \begin{eqnarray*} \label{eq:4} b^{''}&=&2a \left(a_{11}b_{11}+a_{12}b_{21}\right)\left(a_{11}b_{12}+a_{12}b_{22}\right) \\ &+&2c\cdot \left(a_{21}b_{11}+a_{22}b_{21}\right)\left(a_{21}b_{12} +a_{22}b_{22}\right)\\ &+&b\left(\left(a_{11}b_{11} +a_{12}b_{21}\right)\left(a_{21}b_{12} +a_{22}b_{22}\right) +\left(a_{11}b_{12}+a_{12}b_{22}\right)\left(a_{21}b_{11}+a_{22}b_{21}\right)\right) \end{eqnarray*}and by evaluating the factors of $b_{11}b_{12}, b_{21}b_{22}$ , and $b_{11}b_{22}+b_{21}b_{12}$ , it can be checked that \begin{equation*} b^{''} =2a^{'}b_{11}b_{12}+2c^{'}b_{21}b_{22} \\ \nonumber +\left(b_{11}b_{22}+b_{21}b_{12}\right)\left(2aa_{11}a_{12}+2c\cdot a_{21}a_{22}+b\left(a_{11}a_{22}+a_{12}a_{21}\right)\right). \end{equation*}This shows that \begin{equation} [a^{''},b^{''},c^{''}]=[a^{'},b^{'},c^{'}] \ast B \end{equation}and therefore $[a,b,c] \ast (AB)=([a,b,c]\ast A) \ast B$ . Thus, () defines an action of $G$ on the set of (integer) binary quadratic forms. Furthermore, the discriminant of each quadratic form in the orbit of $[a,b,c]$ under $G$ is $b^2-4ac$ .