example of group action

Let $a, b, c$ be integers and let $[a,b,c]$ denote the mapping

[a,b,c]\\colon{\\mathbb{Z}}\\times{\\mathbb{Z}}\\to{\\mathbb{Z}},(x,y)\\mapsto ax^{2}%+bxy+cy^{2}.

Let $G$ be the group of $2\\times 2$ matrices such that $\\det{A}=\\pm 1\\;\\forall\\;A\\in G$ . The substitution

\\left(\\begin{array}[]{l}x\\\\y\\end{array}\\right)\\mapsto A\\left(\\begin{array}[]{l}x\\\\y\\end{array}\\right)

leads to

[a,b,c](a_{11}x+a_{12}y,a_{21}x+a_{22}y)=a^{{}^{\\prime}}x^{2}+b^{{}^{\\prime}}%xy+c^{{}^{\\prime}}y^{2},

where

$\\displaystyle a^{{}^{\\prime}}$	$\\displaystyle=$	$\\displaystyle aa_{11}^{2}+ba_{11}a_{21}+ca_{21}^{2}$	(1)
$\\displaystyle b^{{}^{\\prime}}$	$\\displaystyle=$	$\\displaystyle 2aa_{11}a_{12}+2ca_{21}a_{22}+b(a_{11}a_{22}+a_{12}a_{21})$
$\\displaystyle c^{{}^{\\prime}}$	$\\displaystyle=$	$\\displaystyle aa_{12}^{2}+ba_{12}a_{22}+ca_{22}^{2}$

So we define

[a,b,c]\\ast A:=[a^{{}^{\\prime}},b^{{}^{\\prime}},c^{{}^{\\prime}}]

to be the binary quadratic form with coefficients $a^{{}^{\\prime}},b^{{}^{\\prime}},c^{{}^{\\prime}}$ of $x^{2},xy,y^{2}$ , respectively as in (1). 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

[a,b,c]\\ast(AB)=([a,b,c]\\ast A)\\ast B.

Set $[a,b,c]\\ast(AB):=[a^{{}^{\\prime\\prime}},b^{{}^{\\prime\\prime}},c^{{}^{\\prime%\\prime}}]$ . So we have

$\\displaystyle a^{{}^{\\prime\\prime}}$	$\\displaystyle=$	$\\displaystyle a\\left(a_{11}b_{11}+a_{12}b_{21}\\right)^{2}+c\\left(a_{21}b_{11}+%a_{22}b_{21}\\right)^{2}+b\\left(a_{11}b_{11}+a_{12}b_{21}\\right)(a_{21}b_{11}+a%_{22}b_{21})$	(2)
	$\\displaystyle=$	$\\displaystyle a^{{}^{\\prime}}b_{11}^{2}+c^{{}^{\\prime}}b_{21}^{2}+\\left(2a%\\cdot a_{11}a_{12}+2c\\cdot a_{21}a_{22}+b\\left(a_{11}a_{22}+a_{12}a_{21}\\right%)\\right)b_{11}b_{21}$	(2)
$\\displaystyle c^{{}^{\\prime\\prime}}$	$\\displaystyle=$	$\\displaystyle a\\left(a_{11}b_{12}+a_{12}b_{22}\\right)^{2}+c\\cdot\\left(a_{21}b_%{12}+a_{22}b_{22}\\right)^{2}+b\\cdot\\left(a_{11}b_{12}+a_{12}b_{22}\\right)\\left%(a_{21}b_{12}+a_{22}b_{22}\\right)$	(3)
	$\\displaystyle=$	$\\displaystyle a^{{}^{\\prime}}b_{12}^{2}+c^{{}^{\\prime}}b_{22}^{2}+\\left(2a%\\cdot a_{11}a_{12}+2ca_{21}a_{22}+b\\left(a_{11}a_{22}+a_{12}a_{21}\\right)%\\right)b_{12}b_{22}$	(3)

as desired.For the coefficient $b^{{}^{\\prime\\prime}}$ we get

$\\displaystyle b^{{}^{\\prime\\prime}}$	$\\displaystyle=$	$\\displaystyle 2a\\left(a_{11}b_{11}+a_{12}b_{21}\\right)\\left(a_{11}b_{12}+a_{12%}b_{22}\\right)$
	$\\displaystyle+$	$\\displaystyle 2c\\cdot\\left(a_{21}b_{11}+a_{22}b_{21}\\right)\\left(a_{21}b_{12}+%a_{22}b_{22}\\right)$
	$\\displaystyle+$	$\\displaystyle b\\left(\\left(a_{11}b_{11}+a_{12}b_{21}\\right)\\left(a_{21}b_{12}+%a_{22}b_{22}\\right)+\\left(a_{11}b_{12}+a_{12}b_{22}\\right)\\left(a_{21}b_{11}+a%_{22}b_{21}\\right)\\right)$

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

b^{{}^{\\prime\\prime}}=2a^{{}^{\\prime}}b_{11}b_{12}+2c^{{}^{\\prime}}b_{21}b_{22%}\\\\+\\left(b_{11}b_{22}+b_{21}b_{12}\\right)\\left(2aa_{11}a_{12}+2c\\cdot a_{21}a_{2%2}+b\\left(a_{11}a_{22}+a_{12}a_{21}\\right)\\right).

This shows that

[a^{{}^{\\prime\\prime}},b^{{}^{\\prime\\prime}},c^{{}^{\\prime\\prime}}]=[a^{{}^{%\\prime}},b^{{}^{\\prime}},c^{{}^{\\prime}}]\\ast B

(4)

and therefore $[a,b,c]\\ast(AB)=([a,b,c]\\ast A)\\ast B$ . Thus,(1) defines an action of $G$ on the set of (integer) binaryquadratic forms.Furthermore, the discriminant of each quadratic form in the orbit of $[a,b,c]$ under $G$ is $b^{2}-4ac$ .