Assume that $G$ is a (finite) abelian group and $M$ is a matroid whose ground set is $G$. Let $X$ and $Y$ be subsets of $G$, and $H$ is the stabilizer of $X+Y$. That is $X+Y+H=X+Y$. We denote the rank function of $M$ by $r$. Then can we say that $r(X+Y)\geq$ $r(X)+r(Y)-r(H)$? Or under what condition this can be true? I am fine to assume $M$ to be a partition matroid(or even a uniform matroid) and also assume $G$ is cyclic of prime order.
(A similar result holds for the cardinality of sumsets of a given group, well-known as Kneser's theorem)


Note: To make the statement plausible, adding the condition of $X+Y$ to be independent won't hurt.



Rank of sumsets in matroids
Without any condition on the matroid structure, there is really no reason for your inequality to hold.


For example, take $X=\{a\},Y=\{b\}$ so that $X+Y=\{a+b\}$ where $a,b$ are any nonzero elements of $G$.
Then take any matroid structure such that $a+b$ is a loop but $a,b$ and $0$ are not. We have $0=r(X+Y)<r(X)+r(Y)-r(H)=1$.


Maybe you want to add some kind of compatibility condition between the group and matroid structures to make your question more interesting?


EDIT: Even in a very simple and and well-behaved setting (say $G$ cyclic and $M$ uniform), the inequality fails.
For example, let $G=\mathbb{Z}/N\mathbb{Z}$ with the $n$-uniform matroid structure, where $n<N/2$. Let $X\subset [0, N/2)$ be any subset of size $n$.
Then $r(X+X)=r(X)=r(Y)=n$ and $H$ is trivial so $r(H)=1$. Hence your inequality fails whenever $n>1$.


