Vector space dimension theorem compared to group subsets product formula
First, let's look at what exactly these facts follow from. The second isomorphism theorem for modules and groups imply $$\frac{U+W}{W}\cong \frac{U}{U\cap W}\quad\text{and}\quad \frac{HN}{N}\cong\frac{H}{H\cap N}$$ respectively. Now, applying the fact that dimension and order of groups satisfy: $$\dim(V/W)+\dim(W) = \dim(V)\quad\text{and}\quad |G/K|\cdot|K| = |G|$$ gives you the above equalities. So really you are looking for abstractions of the second isomorphism theorem, and order/dimension.
Universal algebra gives a general framework for discussing the isomorphism theorems. Here are some quick definitions:
- An algebra is a set $A$ equipped with operations on $A$ (maps $\mu:A^k\to A$ for some $k\geq 1$)
- A homomorphism between algebras of the same type is a map of sets $f:A\to B$ which preserves the operations on $A$ and $B$ (see here).
- A congruence $\Phi$ on an alebgra $A$ is an equivalence relation which forms a subalgebra of $A\times A$, where $A\times A$ is given the component-wise algebra structure. From here, we can give the quotient set $A/\Phi$ the same algebra structure, just as we do with groups, rings, modules, etc.
You can find all of this on this page. We then have the following generalization of the second isomorphism theorem:
Let $A$ be an algebra with subalgebra $B$, and let $\Phi$ be a congruence on $A$. Define $$\Phi_B:=\Phi\cap B\times B\quad\text{and}\quad [B]^{\Phi}:=\{X\in A/\Phi\ |\ X\cap B\neq\varnothing\}.$$ Then $\Phi_B$ is a congruence on $B$, $[B]^{\Phi}$ is a subalgebra of $A/\Phi$, and $B/\Phi_B\cong [B]^{\Phi}$.
Now, to get some kind of inclusion-exclusion like you have above, you would need an isomorphism invariant number $|\cdot|$ to associate to the algebra, which is also additive (like vector spaces) or multiplicative (like groups). This would give you the equality $|B/\Phi_B| = |[B]^{\Phi}|$, which generalizes the above two.
Here are two other perspectives, to complement C. Cain's answer.
Connecting the dimension theorem to groups via logarithms:
Vector spaces are, in particular, abelian groups. And finite dimensional vector spaces over finite fields are finite groups. In fact, if $V$ is a vector space over $\mathbb{F}_q$, then $|V| = q^{\dim(V)}$. So in the special case of $\mathbb{F}_q$-vector spaces, the group formula reads $$q^{\dim(V+W)} = \frac{q^{\dim(V)}q^{\dim(W)}}{q^{\dim(V\cap W)}} = q^{\dim(V)+\dim(W) - \dim(V\cap W)}.$$ This is a rather explicit instance of the connection with logs (base $q$, here) that you noticed.
Connecting the dimension theorem to inclusion-exclusion:
There's a general theory of objects supporting notions of dimension and independence, called pregometries by model theorists (like me), and matroids by most other people.
Two canonical examples of pregeometries are
- Vector spaces, in which the closure operator is $\text{cl}(A) = \text{Span}(A)$ and the dimension a set is the linear dimension of its closure. Of course, the dimension theorem $\dim(X\cup Y) = \dim(X) + \dim(Y) - \dim(X\cap Y)$ always holds in a vector space.
- Sets, in which the closure operator is trivial: $\text{cl}(A) = A$ and the dimension is cardinality: $\text{dim}(A) = |A|$. In a set, the dimension theorem is just the inclusion-exclusion principle, so again it always holds.
Not every pregeometry satisfies the dimension theorem - those that do are called modular. A canonical example of a nonmodular pregeometry is a large algebraically closed field, with closure being algebraic closure and dimension being transcendence degree.