# If $p$ is prime, then $\dim(M_k(\Gamma_0(p)) \leq \dim S_k(\Gamma_0(p))+2$

I want to show that $\dim(M_k(\Gamma_0(p)) \leq \dim S_k(\Gamma_0(p))+2$. Where $\dim(M_k(\Gamma_0(p))$ is the Vector Space of holomorphic modular forms on $\Gamma_0(N)=\{\gamma \in SL_2(\mathbb{Z})| \ c \equiv 0\mod N\}$ And $S_k$ the space of Cusp forms. My defintion for these modular forms is that they are invariant for the slash operator for all $\gamma \in \Gamma_0(N)$ and additionally there should be an expansion of the form $$(f|_k\gamma^{-1})(\tau)=\sum_{n=m_\gamma}^\infty a_n q^{n/N}.$$ However, since N is prime here I know that it suffices to check this condition for $\gamma=I$ and $\gamma =S= \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$. My idea was to show that the claim is equivalent to show that that are at most two linearly independent modular forms on $\Gamma_0(p)$ which are no cusp forms. Additionally I know that $M_k(SL_2(\mathbb{Z}))\setminus S_k(SL_2(\mathbb{Z}))=\mathbb{C}G_k$. So I thought that it could be that $$M_k(\Gamma_0(p))\setminus(M_k(SL_2(\mathbb{Z})\cup S_k(\Gamma_0(p))=\mathbb{C}G_k(p\tau).$$

I already know that $G_k(p\tau) \in M_k(\Gamma_0(p))$.

So let $g \in M_k(\Gamma_0(p))\setminus(M_k(SL_2(\mathbb{Z})\cup S_k(\Gamma_0(p)).$ Then I know that it is no cusp form and that

$$g|_kI(\tau)=\sum_{n=0}^\infty a_nq^n$$ Since $T\in \Gamma_0(p)$. And $$g|_kS(\tau)=\sum_{n=0}^\infty b_nq^{n/p}$$ I think that I have to show $g-\lambda*G_k(p\tau)\in S_k(\Gamma_0(p))$ for $\lambda \in \mathbb{C}$. However I do not know if this is even a right approach, since if $M_k(\Gamma_0(p))\setminus S_k(\Gamma_0(p))=\mathbb{C}(G_k(\tau)\oplus G_k(p\tau))$. Then the inequality that I want to show had to be always an equality. I am grateful for any help :).