[proofplan] We represent an arbitrary point of $\mathbb{P}^1(\mathbb{Q})$ by a primitive integer vector $(a,c) \in \mathbb{Z}^2$. The coprimality condition gives integers completing $(a,c)$ to a unimodular matrix. That matrix lies in $SL_2(\mathbb{Z})$ and sends $\infty=[1:0]$ to $[a:c]$, proving that every rational projective point is in the same orbit. [/proofplan] [step:Represent the projective point by a primitive integer vector] Let $p \in \mathbb{P}^1(\mathbb{Q})$ be arbitrary. Choose homogeneous coordinates \begin{align*} p = [r:s] \end{align*} with $r,s \in \mathbb{Q}$ not both zero. Multiplying both coordinates by a common nonzero integer, choose $a,c \in \mathbb{Z}$ not both zero such that \begin{align*} p = [a:c]. \end{align*} Dividing $a$ and $c$ by their positive greatest common divisor, we may assume \begin{align*} \gcd(a,c)=1. \end{align*} Thus every point of $\mathbb{P}^1(\mathbb{Q})$ has a representative $[a:c]$ with $(a,c) \in \mathbb{Z}^2$ primitive. [/step] [step:Complete the primitive vector to a unimodular matrix] By Bezout's identity applied to the coprime integers $a$ and $c$, there exist integers $u,v \in \mathbb{Z}$ such that \begin{align*} au + cv = 1. \end{align*} Define $d := u$ and $b := -v$. Then \begin{align*} ad - bc = au - b c = au + cv = 1. \end{align*} Therefore the matrix \begin{align*} M := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align*} has integer entries and determinant $1$, so $M \in SL_2(\mathbb{Z})$. [guided] The goal is to find an element of $SL_2(\mathbb{Z})$ whose first column is the primitive vector $(a,c)$. The determinant condition for a matrix \begin{align*} M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align*} is precisely \begin{align*} \det M = ad - bc. \end{align*} Because $\gcd(a,c)=1$, Bezout's identity applies to the pair of integers $a$ and $c$. It states that the integer linear combinations of $a$ and $c$ contain $1$, so there exist $u,v \in \mathbb{Z}$ satisfying \begin{align*} au + cv = 1. \end{align*} Set $d := u$ and $b := -v$. Then \begin{align*} ad - bc = au - (-v)c = au + cv = 1. \end{align*} Thus $M$ has integer entries and determinant $1$, so $M \in SL_2(\mathbb{Z})$. [/guided] [/step] [step:Send $\infty$ to the chosen rational projective point] Let $\infty := [1:0] \in \mathbb{P}^1(\mathbb{Q})$. By the defining action of $SL_2(\mathbb{Z})$ on homogeneous coordinates, \begin{align*} M \cdot \infty &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot [1:0] \\ &= [a\cdot 1 + b\cdot 0 : c\cdot 1 + d\cdot 0] \\ &= [a:c] \\ &= p. \end{align*} Since $p \in \mathbb{P}^1(\mathbb{Q})$ was arbitrary, every point of $\mathbb{P}^1(\mathbb{Q})$ lies in the $SL_2(\mathbb{Z})$-orbit of $\infty$. Hence the action is transitive. [/step] [step:Identify the single cusp] By definition, the cusps of the full modular group are the orbits of $SL_2(\mathbb{Z})$ on $\mathbb{P}^1(\mathbb{Q})$. The transitivity just proved says that there is exactly one such orbit, namely \begin{align*} SL_2(\mathbb{Z}) \cdot \infty = \mathbb{P}^1(\mathbb{Q}). \end{align*} Therefore $SL_2(\mathbb{Z}) \backslash \mathbb{P}^1(\mathbb{Q})$ has one element, and the full modular group has exactly one cusp. [/step]