TITLE: Category-Theoretic relation between Orbit-Stabilizer and Rank-Nullity Theorems
QUESTION [11 upvotes]: In linear algebra, the Rank-Nullity theorem states that given a vector space $V$ and an $n\times n$ matrix $A$,
$$\text{rank}(A) + \text{null}(A) = n$$
or that
$$\text{dim(image}(A)) + \text{dim(ker}(A)) = \text{dim}(V).$$

In abstract algebra, the Orbit-Stabilizer theorem states that given a group $G$ of order $n$, and an element $x$ of the set $G$ acts on,
$$|\text{orb}(x)||\text{stab}(x)| = |G|.$$

Other than the visual similarity of the expressions, is there some deeper, perhaps category-theoretic connection between these two theorems? Is there, perhaps, a functor from the category of groups $\text{Grp}$ to some category where linear transformations are morphisms? Am I even using the words functor and morphism correctly in this context?

REPLY [3 votes]: The intuition behind this question is spot-on. I'm going to try to fill out some of the details to make this work.
The first thing to note is that a linear map $A:V\to V$ also gives a genuine group action: it is the additive group of $V$ acting on the set $V$ by addition. That is, any $v\in V$ acts on $x\in V$ as $v: x \mapsto x+Av.$ 
Now we see that given any $x$ in $V$ the stabilizer subgroup $\text{stab}(x)$ of this action is precisely the kernel of $A.$ The orbit of $x$ is $x$ plus the image of $A.$
If we are working with a vector space over a finite field, we can take the cardinality of these sets as in the formula $|\text{orb}(x)||\text{stab}(x)| = |G|$ and as @Ravi suggests, take the logarithm of this where the base is the size of the field and we get exactly the rank-nullity equation.
If we have an infinite field then this doesn't work and we need to think more along the lines of a categorified orbit-stabilizer theorem. In this case, for each $x\in V$ we can find a bijection:
$$
\text{orb}(x) \cong G / \text{stab}(x)
$$
and as @Nick points out, this bijection gives us the First Isomorphism Theorem:
$$
\mathrm{Im}(A) \cong V / \mathrm{Ker}(A).
$$