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Use the three left-coset representatives $I,S,ST$ for ${\Gamma }_0(2)\backslash {\mathrm{SL}}_2(\mathbb{Z})$. If a nonzero weight-four cusp form $h$ existed, its coset norm of a modular formwould be a nonzero weight-twelve modular form for the full group: right multiplication permutes its factors. At infinity $h$ has order at least one in $q$. The other two factors correspond to the zero modular cusp of cusp width two and each have order at least $1/2$ in $q$. Thus $H$ has order at least two there.

The ratio $H/\Delta$ is weight zero, holomorphic on the half-plane because $\Delta$ has no zeros there, and holomorphic at the modular cusp with value zero. It descends to a holomorphic function on the compact full modular curve. Such a function is constant, hence zero, contradicting $H\ne 0$. ThereforeConstant terms at the two modular cusps define a linear map $M_4({\Gamma }_0(2))\to {\mathbb{C}}^2$ whose kernel is this modular cusp space. It is injective, so the dimension is at most two. The forms $E_4(\tau )$ and $E_4(2\tau )$ are holomorphic weight-four forms for this group. For the second, the same conjugation used in 1(c) proves transformation, and $E_4(2S\tau )|$ after the weight factor is $2^{-4}{E}_4(\tau /2)$, proving holomorphy at zero. Their constant terms are both one but their $q$ coefficients are respectively $240$ and zero, so they are independent. We obtainThis is the weight-four Eisenstein basis at level two.

There are no weight-four cusp forms at level two. A nonzero form would have a weight-twelve coset norm of a modular form with modular cusp order at least two, contradicting the simple modular cusp order and interior nonvanishing of the modular discriminant. The two independent displayed Eisenstein series span the weight-four space, since its cusp-constant map embeds it in a two-dimensional space.