An abelian subgroup of symmetric group PROBLEM: Let $G$ be an abelian subgroup of the symmetric group $S_n$ and $p_1, . . . , p_k$ be all prime divisors of $|G|$. Prove that $n≥p_1 +···+p_k$.
QUESTION: How do you solve this problem. I've thought about using the Sylow theorems but I can't seem to get the final inequality.

I am assuming these are the distinct prime divisors of $|G|$. In that case $G$ has an element of order $p_1p_2\cdots p_k$ since it is abelian. The order of an element of $S_n$ is the least common multiple of the lengths of its disjoint cycles. These lengths must sum to a number less than or equal to $n$. Since the $p_i$ are prime, there must be at least one cycle length divisible by each of the primes (with some cycle lengths possibly divisible by multiple primes). Since $xy\geq x+y$ whenever $x,y\geq 2$, the smallest possible sum of the lengths of the cycles in the element with this order is $p_1+p_2+\cdots+p_k\leq n$.
