[proofplan] The central point is that the matrix $-I_2$ belongs to $SL_2(\mathbb{Z})$ but fixes every point of the upper half-plane under the fractional linear action. The modular transformation law therefore forces a weight factor of $(-1)^k$ without changing the input point. When $k$ is odd this factor is $-1$, so every modular form must equal its negative at every point of $\mathbb{H}$. [/proofplan] [step:Apply the modularity condition to the central matrix $-I_2$] Let \begin{align*} \gamma_0 := -I_2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}. \end{align*} Then $\gamma_0 \in SL_2(\mathbb{Z})$, since its entries are integers and \begin{align*} \det(\gamma_0) = (-1)(-1) - 0 \cdot 0 = 1. \end{align*} Let \begin{align*} f: \mathbb{H} &\to \mathbb{C} \end{align*} be an element of $M_k(SL_2(\mathbb{Z}))$. For a matrix \begin{align*} \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}), \end{align*} the weight $k$ slash transform of $f$ by $\gamma$ is the function \begin{align*} f|_k \gamma: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto (cz+d)^{-k} f\left(\frac{az+b}{cz+d}\right). \end{align*} Since $f$ is a modular form of weight $k$ for $SL_2(\mathbb{Z})$, it satisfies \begin{align*} f|_k \gamma = f \end{align*} for every $\gamma \in SL_2(\mathbb{Z})$. In particular, \begin{align*} f|_k \gamma_0 = f. \end{align*} [/step] [step:Compute the action of $-I_2$ on the upper half-plane] For $z \in \mathbb{H}$, the fractional linear action of $\gamma_0$ gives \begin{align*} \frac{(-1)z + 0}{0z + (-1)} = z. \end{align*} Also, the automorphy factor for $\gamma_0$ is \begin{align*} cz+d = 0z - 1 = -1. \end{align*} Therefore, for every $z \in \mathbb{H}$, \begin{align*} (f|_k \gamma_0)(z) &= (-1)^{-k} f(z). \end{align*} Because $k \in \mathbb{Z}$, we have $(-1)^{-k} = (-1)^k$. Hence \begin{align*} (f|_k \gamma_0)(z) &= (-1)^k f(z). \end{align*} [/step] [step:Use oddness of the weight to force the modular form to vanish] Since $k$ is odd, \begin{align*} (-1)^k = -1. \end{align*} Combining this with $f|_k \gamma_0 = f$, we obtain, for every $z \in \mathbb{H}$, \begin{align*} f(z) &= (f|_k \gamma_0)(z) \\ &= -f(z). \end{align*} Thus \begin{align*} 2f(z) = 0. \end{align*} Since the codomain of $f$ is $\mathbb{C}$, whose characteristic is zero, it follows that $f(z)=0$ for every $z \in \mathbb{H}$. Therefore $f$ is the zero function, and since $f \in M_k(SL_2(\mathbb{Z}))$ was arbitrary, \begin{align*} M_k(SL_2(\mathbb{Z})) = \{0\}. \end{align*} [/step] [step:Deduce the vanishing of cusp forms] By definition, every cusp form of weight $k$ for $SL_2(\mathbb{Z})$ is in particular a modular form of weight $k$ for $SL_2(\mathbb{Z})$, so \begin{align*} S_k(SL_2(\mathbb{Z})) \subseteq M_k(SL_2(\mathbb{Z})). \end{align*} Since $M_k(SL_2(\mathbb{Z})) = \{0\}$, this inclusion gives \begin{align*} S_k(SL_2(\mathbb{Z})) = \{0\}. \end{align*} [/step]