Does $a!b!$ always divide $(a+b)!$ Hello the question is as stated above and is given to us in the context of group theory, specifically under the heading of isomorphism and products. I would write down what I have tried so far but I have made very little progress in trying to solve this over the last few hours! 

A proof based on the binomial recurrence: $\frac{(a+b)!}{a!b!} ( =\binom{a+b}{a}) =: f(a,b)$ fulfills the following recurrence:
$f(a,b) = f(a-1,b) + f(a,b-1), \quad a>0,b>0$
with starting values
$f(a,0) = f(0,a) = 1, \quad a \ge 0$
so by induction for all non-negative integer values $a,b$ the function $f(a,b)=\frac{(a+b)!}{a!b!}$ is an integer value.
