[proofplan] We first record that the weight-$k$ slash operators form a right action of $SL_2(\mathbb{Z})$ on functions $\mathbb{H} \to \mathbb{C}$. We then prove directly, using the Euclidean algorithm on the first column, that $S$ and $T$ generate $SL_2(\mathbb{Z})$. Finally, invariance under $S$ and $T$ gives invariance under their inverses, and the right-action identity propagates the invariance through any word in the generators. [/proofplan]

[step:Verify that the slash operators compose as a right action]Let \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad \delta = \begin{pmatrix} e & r \\ g & h \end{pmatrix} \end{align*} be elements of $SL_2(\mathbb{Z})$. Define \begin{align*} \gamma z := \frac{az+b}{cz+d}, \qquad \delta z := \frac{ez+r}{gz+h}, \qquad z \in \mathbb{H}. \end{align*} The standard fractional linear action of $SL_2(\mathbb{Z})$ preserves $\mathbb{H}$, so all displayed expressions are defined for $z \in \mathbb{H}$. The product is \begin{align*} \gamma\delta = \begin{pmatrix} ae+bg & ar+bh \\ ce+dg & cr+dh \end{pmatrix}. \end{align*} For $z \in \mathbb{H}$, \begin{align*} ((f|_k\gamma)|_k\delta)(z) &= (gz+h)^{-k}(f|_k\gamma)(\delta z) \\ &= (gz+h)^{-k}\left(c\frac{ez+r}{gz+h}+d\right)^{-k} f\left(\gamma(\delta z)\right) \\ &= \bigl((ce+dg)z+(cr+dh)\bigr)^{-k} f\left(\frac{(ae+bg)z+(ar+bh)}{(ce+dg)z+(cr+dh)}\right) \\ &= (f|_k(\gamma\delta))(z). \end{align*} Thus \begin{align*} (f|_k\gamma)|_k\delta = f|_k(\gamma\delta) \end{align*} for all $\gamma,\delta \in SL_2(\mathbb{Z})$.[/step]

[guided]The first structural fact we need is that the slash operation is compatible with matrix multiplication. Let \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad \delta = \begin{pmatrix} e & r \\ g & h \end{pmatrix} \end{align*} be matrices in $SL_2(\mathbb{Z})$. We write their fractional linear actions as maps \begin{align*} \gamma: \mathbb{H} &\to \mathbb{H}, & z &\mapsto \frac{az+b}{cz+d}, \\ \delta: \mathbb{H} &\to \mathbb{H}, & z &\mapsto \frac{ez+r}{gz+h}. \end{align*} These maps are defined on $\mathbb{H}$ because matrices in $SL_2(\mathbb{R})$ act on the upper half-plane by fractional linear transformations. Now compute the iterated slash operation. For $z \in \mathbb{H}$, \begin{align*} ((f|_k\gamma)|_k\delta)(z) &= (gz+h)^{-k}(f|_k\gamma)(\delta z) \\ &= (gz+h)^{-k} \left(c\frac{ez+r}{gz+h}+d\right)^{-k} f\left(\gamma(\delta z)\right). \end{align*} The automorphy factors multiply as \begin{align*} (gz+h)\left(c\frac{ez+r}{gz+h}+d\right) &= c(ez+r)+d(gz+h) \\ &= (ce+dg)z+(cr+dh). \end{align*} Also, \begin{align*} \gamma(\delta z) &= \frac{a\frac{ez+r}{gz+h}+b}{c\frac{ez+r}{gz+h}+d} \\ &= \frac{(ae+bg)z+(ar+bh)}{(ce+dg)z+(cr+dh)}. \end{align*} Since \begin{align*} \gamma\delta = \begin{pmatrix} ae+bg & ar+bh \\ ce+dg & cr+dh \end{pmatrix}, \end{align*} the last two displayed identities give \begin{align*} ((f|_k\gamma)|_k\delta)(z) = (f|_k(\gamma\delta))(z). \end{align*} Because this holds for every $z \in \mathbb{H}$, we have \begin{align*} (f|_k\gamma)|_k\delta = f|_k(\gamma\delta). \end{align*} This is the right-action identity used below.[/guided]

[step:Generate every matrix in $SL_2(\mathbb{Z})$ from $S$ and $T$]Let $G$ be the subgroup of $SL_2(\mathbb{Z})$ generated by $S$ and $T$. We prove that $G = SL_2(\mathbb{Z})$. Let \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}). \end{align*} Since $\det \gamma = ad-bc = 1$, the integers $a$ and $c$ are coprime. We shall left-multiply $\gamma$ by elements of $G$ until the lower-left entry becomes $0$. If $c=0$, then $ad=1$, so $a=d=\varepsilon$ for some $\varepsilon \in \{1,-1\}$, and \begin{align*} \gamma = \begin{pmatrix} \varepsilon & b \\ 0 & \varepsilon \end{pmatrix} = \varepsilon I\, T^{\varepsilon b}. \end{align*} Since $S^2=-I$, both $I$ and $-I$ belong to $G$, and therefore $\gamma \in G$. Assume now that $c \neq 0$. If $|a| < |c|$, left-multiplication by $S$ replaces the first column $(a,c)^\top$ by $(-c,a)^\top$, so the absolute value of the lower entry becomes $|a|$. If $|a| \ge |c|$, choose $q \in \mathbb{Z}$ such that the integer $r:=a+qc$ satisfies $|r| < |c|$; this is the [division algorithm](/theorems/725) applied to $a$ by $c$. Left-multiplication by $T^q$ replaces the first column $(a,c)^\top$ by $(a+qc,c)^\top=(r,c)^\top$, and then left-multiplication by $S$ replaces it by $(-c,r)^\top$. In either case, the absolute value of the lower entry strictly decreases. Repeating this finite descent in the nonnegative integer $|c|$, we obtain an element $A \in G$ such that \begin{align*} A\gamma = \begin{pmatrix} \varepsilon & m \\ 0 & \varepsilon \end{pmatrix} \end{align*} for some $\varepsilon \in \{1,-1\}$ and $m \in \mathbb{Z}$. By the already treated case, $A\gamma \in G$. Since $A \in G$ and $G$ is a subgroup, $\gamma=A^{-1}(A\gamma)\in G$. Hence $SL_2(\mathbb{Z}) \subseteq G$, and the reverse inclusion holds by definition of $G$. Therefore $SL_2(\mathbb{Z})=G$.[/step]

[guided]We need the concrete group-theoretic input that $S$ and $T$ generate $SL_2(\mathbb{Z})$. Let $G$ denote the subgroup of $SL_2(\mathbb{Z})$ generated by $S$ and $T$. We prove $G=SL_2(\mathbb{Z})$ by reducing an arbitrary matrix using the Euclidean algorithm. Take \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}). \end{align*} The determinant condition gives \begin{align*} ad-bc=1, \end{align*} so the first-column entries $a$ and $c$ have greatest common divisor $1$. We now act on the left by powers of $T$ and by $S$. Left multiplication changes rows, hence changes the first column as follows: \begin{align*} T^q \begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} a+qc & b+qd \\ c & d \end{pmatrix}, \\ S \begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} -c & -d \\ a & b \end{pmatrix}. \end{align*} Thus $T^q$ replaces the first column $(a,c)^\top$ by $(a+qc,c)^\top$, while $S$ replaces it by $(-c,a)^\top$. The goal is to make the lower-left entry equal to $0$. If $c=0$, then $ad=1$, so $a=d=\varepsilon$ for some $\varepsilon \in \{1,-1\}$. Therefore \begin{align*} \gamma = \begin{pmatrix} \varepsilon & b \\ 0 & \varepsilon \end{pmatrix} = \varepsilon I\, T^{\varepsilon b}. \end{align*} Since $S^2=-I$, the matrices $I$ and $-I$ both lie in $G$, and $T^{\varepsilon b}\in G$. Hence $\gamma\in G$ in this case. Now suppose $c\neq 0$. We show how to reduce $|c|$. If $|a|<|c|$, then left-multiplication by $S$ sends the first column to $(-c,a)^\top$, so the new lower-left entry has absolute value $|a|<|c|$. If $|a|\ge |c|$, the division algorithm gives an integer $q$ such that \begin{align*} r:=a+qc \end{align*} satisfies $|r|<|c|$. First left-multiply by $T^q$, which changes the first column to $(r,c)^\top$. Then left-multiply by $S$, which changes it to $(-c,r)^\top$. The new lower-left entry is $r$, and therefore its absolute value is strictly smaller than $|c|$. This gives a strict descent in the nonnegative integer $|c|$. After finitely many repetitions, the lower-left entry becomes $0$. Hence there exists $A\in G$ such that \begin{align*} A\gamma = \begin{pmatrix} \varepsilon & m \\ 0 & \varepsilon \end{pmatrix} \end{align*} for some $\varepsilon\in\{1,-1\}$ and $m\in\mathbb{Z}$. The triangular case already proves $A\gamma\in G$. Since $A\in G$ and $G$ is a subgroup, $A^{-1}\in G$, and therefore \begin{align*} \gamma=A^{-1}(A\gamma)\in G. \end{align*} Thus every element of $SL_2(\mathbb{Z})$ lies in $G$, so $SL_2(\mathbb{Z})=G$.[/guided]

[step:Deduce invariance under inverse generators] The identity matrix $I \in SL_2(\mathbb{Z})$ satisfies $f|_k I=f$ by the definition of the slash operator. Since $SS^{-1}=I$, the right-action identity gives \begin{align*} f = f|_k I = f|_k(SS^{-1}) = (f|_k S)|_k S^{-1} = f|_k S^{-1}. \end{align*} Similarly, since $TT^{-1}=I$, \begin{align*} f = f|_k I = f|_k(TT^{-1}) = (f|_k T)|_k T^{-1} = f|_k T^{-1}. \end{align*} Thus $f$ is invariant under $S$, $S^{-1}$, $T$, and $T^{-1}$. [/step]

[step:Propagate invariance through a word for $\gamma$]Let $\gamma \in SL_2(\mathbb{Z})$. Since $S$ and $T$ generate $SL_2(\mathbb{Z})$, there exist an integer $N \ge 0$ and matrices \begin{align*} g_1,\dots,g_N \in \{S,S^{-1},T,T^{-1}\} \end{align*} such that \begin{align*} \gamma = g_1g_2\cdots g_N, \end{align*} with the convention that $N=0$ means $\gamma=I$. We prove by induction on $j \in \{0,1,\dots,N\}$ that \begin{align*} f|_k(g_1g_2\cdots g_j)=f, \end{align*} where the product for $j=0$ is $I$. The case $j=0$ is $f|_k I=f$. Suppose the assertion holds for some $j<N$. Using the right-action identity and the fact that $f|_k g_{j+1}=f$, we obtain \begin{align*} f|_k(g_1\cdots g_jg_{j+1}) &= (f|_k(g_1\cdots g_j))|_k g_{j+1} \\ &= f|_k g_{j+1} \\ &= f. \end{align*} By induction, $f|_k(g_1\cdots g_N)=f$. Since $g_1\cdots g_N=\gamma$, this gives \begin{align*} f|_k\gamma=f. \end{align*} As $\gamma \in SL_2(\mathbb{Z})$ was arbitrary, the theorem follows.[/step]

[guided]Fix an arbitrary matrix $\gamma \in SL_2(\mathbb{Z})$. The generation result says that $\gamma$ can be written as a finite word in the four matrices $S$, $S^{-1}$, $T$, and $T^{-1}$. Thus there are an integer $N\ge 0$ and matrices \begin{align*} g_1,\dots,g_N \in \{S,S^{-1},T,T^{-1}\} \end{align*} such that \begin{align*} \gamma=g_1g_2\cdots g_N. \end{align*} If $N=0$, this means $\gamma=I$. We now move through this word one letter at a time. Define, for each $j\in\{0,1,\dots,N\}$, the partial product \begin{align*} \gamma_j := \begin{cases} I, & j=0,\\ g_1g_2\cdots g_j, & 1\le j\le N. \end{cases} \end{align*} We prove by induction on $j$ that \begin{align*} f|_k\gamma_j=f. \end{align*} For $j=0$, this is $f|_k I=f$, which follows directly from the definition of the slash operator. Assume $f|_k\gamma_j=f$ for some $j<N$. Since $g_{j+1}$ is one of $S$, $S^{-1}$, $T$, or $T^{-1}$, the previous step gives \begin{align*} f|_k g_{j+1}=f. \end{align*} Using the right-action identity, \begin{align*} f|_k\gamma_{j+1} &= f|_k(\gamma_j g_{j+1}) \\ &= (f|_k\gamma_j)|_k g_{j+1} \\ &= f|_k g_{j+1} \\ &= f. \end{align*} This proves the induction step. Taking $j=N$, we get \begin{align*} f|_k\gamma_N=f. \end{align*} Since $\gamma_N=\gamma$, this is exactly \begin{align*} f|_k\gamma=f. \end{align*} Because the original choice of $\gamma \in SL_2(\mathbb{Z})$ was arbitrary, $f$ is invariant under every element of $SL_2(\mathbb{Z})$.[/guided]