Using group theory, show that $\forall \ n\in \mathbb Z_{\geq 0}, \ (n+1)$ divides ${2n}\choose{n}$

Question

That $(n+1)$ divides ${2n}\choose{n}$ can be proven in different ways as done here.

$$\frac{1}{n+1}\binom{2n}{n} = \binom{2n}{n} - \binom{2n}{n+1}$$

Every Catalan number $C_n=\frac{1}{n+1}\binom{2n}{n}$ is certainly a non-negative integer (because it's the number of possible dyck words of length $2n$).

However, my professor asked me to prove this using a group theory argument. I am not sure if this is even possible.

I have thought of Lagrange's theorem which states that:

For any finite group say $G$, the order of a subgroup $H$ of group $G$ divides the order of $G$.

But I can't think of any possible situation where a group of order $\binom{2n}{n}$ has a subgroup of order $n+1$.

How often is group theory used to prove divisibility problems? I'm new to Abstract Algebra.

Answer 1

There are several proofs showing that $C_n$ is an integer, i.e., that $n+1$ divides $\binom{2n}{n}$, by using groups and algebras. Here are a few of them.

$1.$ Let $A$ be the subalgebra of $M_{n-1}(K)$ consisting of upper-triangular matrices of size $n-1$. Then the number of two-sides ideals in $A$ equals $C_n$, i.e., $n+1$ divides $\binom{2n}{n}$.

$2.$ The number of permutations in the group $S_n$ which have no increasing subsequence of length $3$ is the Catalan number $C_n$. Hence $n+1$ divides $\binom{2n}{n}$.

$3.$ Let the symmetric group $S_n$ act on the polynomial ring $A = \Bbb C[x_1,\ldots,x_n,y_1,...,y_n]$ by $$ w · f (x_1, \ldots , x_n, y_1, . . . , y_n) = f (x_{w(1)}, \ldots , x_{w(n)}, y_{w(1)}, \ldots , y_{w(n)}) $$ for all $w ∈ S_n$. Let $I$ be the ideal generated by all invariants of positive degree, i.e., $$ I = ⟨f ∈ A \mid w · f = f \text{ for all } w ∈ S_n \text{ and }f(0) = 0⟩. $$ Then $C_n$ is the dimension of the subspace $U$ of $A/I$ given by $$ U=\{f ∈ A/I \mid w · f = ({\rm sgn} ( w) )f \text{ for all }w ∈ S_n \}. $$ Hence $C_n$ is an integer and $n+1$ divides $\binom{2n}{n}$.

$4.$ In the theory of algebraic curves, the Catalan number $C_n$ counts the covers $C → \Bbb P^1$ of minimal degree $n + 1$ from a general curve $C$ of genus $2n$ . Each such cover has simple ramification and its monodromy group equals the symmetric group $S_{n+1}$. The number of such covers coincides with the degree of the Grassmannian $G (2, n + 2)$ in its Plücker embedding, which is well-known to equal $C_n$ .