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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.

Nothing special
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  • It's not that obvious that $n+1$ divides the binomial coefficient... – lulu Apr 09 '23 at 12:31
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    "How often is group theory used to prove divisibility problems?" It is used often. Also the converse is used, i.e., group theory problems are often reduced to divisibility in the ring $\Bbb Z$. – Dietrich Burde Apr 09 '23 at 12:34
  • But certainly, the usual proofs are not using group theory here, compare with this post. – Dietrich Burde Apr 09 '23 at 12:36
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    Still not obvious. Are you thinking that $m>n+1$ implies that $\binom mn$ is divisible by $n+1$? That's just not true. $\binom 53=10$ which is not divisible by $4$. – lulu Apr 09 '23 at 12:36
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    here are some of the standard proofs via binomial coefficients. I wouldn't call any of them "obvious". – lulu Apr 09 '23 at 12:37
  • My bad! I'm editing it. I got it wrong... – Nothing special Apr 09 '23 at 12:40
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    As to the underlying question, it's not obvious to me what argument your professor has in mind. There are versions of the Catalan numbers that are attached to each of the Coxeter groups...the standard Catalan numbers are attached to $A_n$ for example. See this. Not sure that anyone would be expected to come up with this entirely on their own, however. – lulu Apr 09 '23 at 12:47
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    Arguably the sixth proof on the Wikipedia page is a group theory proof. However, 1) it doesn't use Lagrange's theorem -- it uses the more general fact that if a group acts freely on a set, then the order of the group divides the size of the set, and 2) the group in question is $\mathbb{Z}/(2n+1)$: it is a proof that $2n+1$ divides $\binom{2n+1}{n}$ rather than a proof that $n+1$ divides $\binom{2n}{n}$. So I'm not sure it's exactly what you're looking for... – Micah Apr 09 '23 at 13:29

1 Answers1

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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 $C_n$ .

Dietrich Burde
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