Suppose we are in vacuum and we have an electric field $\vec{E}$ which we assume is simple harmonic wave that propagates through $z$ and is linearly polarized in the $x$-$y$ plane along $x$ i.e. $\vec{E}(t,x,y,z)=E_0\cos(\omega t-kz)\hat{x}$. This function obviously satisfies $\vec{\nabla}\cdot\vec{E}=0$. Now note that $$\vec{\nabla}\times\vec{E}(t,x,y,z)=\frac{\partial}{\partial z}\left( E_0\cos(\omega t-kz) \right)\hat{y}=kE_0\sin(\omega t-kz)\hat{y}=-\frac{\partial \vec{B}}{\partial t}(t,x,y,z)$$ If we integrate this, we get $\vec{B}(t,x,y,z)=B_0\cos(\omega t-kz)\hat{y}+\vec{g}(x,y,z)$ for some $\vec{g}:\mathbb{R}^3\rightarrow\mathbb{R}^3$ and where $B_0=\frac{k}{\omega}E_0$. Now note that $B_0\cos(\omega t-kz)\hat{y}$ has no gradient and therefore $\vec{\nabla}\cdot\vec{B}=0$ implies $\vec{\nabla}\cdot\vec{g}=0$. On the other hand $\vec{\nabla}\times (B_0\cos(\omega t-kz)\hat{y})=\mu_0\epsilon_0\frac{\partial\vec{E}}{\partial t}(t,x,y,z)$, which implies $\vec{\nabla}\times\vec{g}=\vec{0}$. Finally, clearly $B_0\cos(\omega t-kz)\hat{y}$ satisfies the wave equation, which means that $\vec{g}$ must do it too. Since $\vec{g}$ has no time dependence, $\nabla^2\vec{g}=\vec{0}$.
On every source I have seen, the vector field $\vec{g}$ has been taken to be null. From this assumption things such as the perpendicularity between the fields and the direction of propagation are explained. Non the less, non of the three restrictions imply that $\vec{g}$ be null. In particular, $\vec{g}$ could be a constant vector field with any direction and still satisfy Maxwell's equations.
Is there any way to show that in general E&M waves must be perpendicular and therefore show that $\vec{g}=\vec{0}$? In the case there isn't, what can we say about $\vec{g}$ knowing $\vec{\nabla}\cdot\vec{g}=0$, $\vec{\nabla}\times\vec{g}=\vec{0}$ and $\nabla^2\vec{g}=\vec{0}$?
