Let $\mathbb{P}$ be a forcing. If $G \subseteq \mathbb{P}$ and $H \subseteq \mathbb{P}$ are two $\mathbb{P}$-generic filters over $V$ and $G \neq H$, does this imply that $M[G] \neq M[H]$.
If $G$ and $H$ are mutually generic, this should be true; however, what about in general?
It's not necessarily true in general. Suppose that $\pi\in M$ is an automorphism of $\Bbb P$, then $H=\pi''G$ is a generic filter as well. And clearly, $G\in M[H]$ and $H\in M[G]$, and therefore $M[H]=M[G]$.
If $\pi$ is such automorphism that for some $p\in G$ we have that $p$ and $\pi p$ are incompatible then $H\neq G$, since $\pi p\in H$ and $p\in G$. One nice example for such situation are the Cohen forcing, consider $\Bbb P$ to be the set of all finite functions from $\omega$ to $2$, and consider the automorphism defined by: $$\pi p(n)=\begin{cases}1-p(n) & n=0\\ p(n)&\text{otherwise}\end{cases}$$