Proving that $a!b!$ divides $\left(a + b \right)!$ using group theory

Question

I was wondering whether this argument proves that $a!b!$ divides $\left(a + b \right)!$, and whether one could make it more rigorous.

Let $S_{a+b}$ be the symmetry group of $\{1, 2, \ldots , a+b\}$. Describe a subgroup of $S_{a+b}$ isomorphic to $S_a\times S_b$ as the set of permutations that permute $\{1,\ldots,a\}$ amongst themselves, and $\{a+1,\ldots,a+b\}$ amongst themselves. Then, by Lagrange's Theorem, $|S_a\times S_b|=|S_a||S_b| = a!b!$ divides $|S_{a+b}|=(a+b)!$.

Answer 1

You've absolutely got the right idea! To make it more rigorous, I would proceed as follows:

• Let $H$ be the set of $\sigma\in S_{a+b}$ such that whenever $n\in\Bbb N$ such that $a<n\le a+b,$ we have $\sigma(n)=n.$
• Let $K$ be the set of $\sigma\in S_{a+b}$ such that whenever $n\in\Bbb N$ such that $n<a+1,$ we have $\sigma(n)=n.$
• Demonstrate isomorphisms $f:S_a\to H$ and $g:S_b\to K.$
• Let $G=HK:=\{hk:h\in H,k\in K\},$ and show that for all $h\in H,k\in K,$ we have $hk=kh.$ Thus, $G$ is a subgroup of $S_{a+b},$ and the map $H\times K\to G$ given by $(h,k)\mapsto hk$ is a homomorphism.
• Show that $H\cap K=\{\text{id}\},$ and so the aforementioned map $H\times K\to G$ is injective, so surjective, and so an isomorphism.
• Show that the map $S_a\times S_b\to G$ given by $(\rho,\sigma)\mapsto f(\rho)g(\sigma)$ is an isomorphism, and so we can use Lagrange's Theorem, as desired.

Let me know if you have difficulty with any of these steps, or if you just want to bounce your attempts off of someone.