Proof containing abelian groups. Let $G \leq S_{999}$ be an abelian subgroup of order $|G| = 1111$. Prove that there exists $i \in$ {$1,2,...,999$} such that $\forall α \in G, α(i) = i$.
Okay so I came across this problem and even though it looked easy in the beginning, it started to confuse me. It looks like a tricky proof and I don't know where to begin. Any ideas?
 A: The order of $G$, $1111=11\cdot 101$, is the product of two primes.  Since it is abelian, it must be cyclic.  Let $\sigma$ be a generator.
Since $1111 > 999$, $\sigma$ must be a product of some $11$-cycles and some $101$-cycles.  But $999$ cannot be written as a sum of nonnegative multiples of $11$ and $101$, so $\sigma$, and therefore $G$, must have a fixed point.

I suspect that the motivation for this problem is the coin problem, specifically the $n=2$ case, due to Sylvester in 1884.
The result is that, if $a$ and $b$ are relatively prime positive integers, then every integer $\geq (a-1)(b-1) = ab-a-b+1$ can be written as a sum of multiples of $a$ and multiples of $b$, while $ab-a-b$ cannot.
To see that we cannot have $ab-a-b=ak+bl$, where $k$ and $l$ are non-negative integers, consider the equation modulo $a$ and $b$.  This gives us $l\equiv -1\pmod{a}$ and $k\equiv -1\pmod{b}$.  But then $ak+bl \geq a(b-1) + b(a-1) = 2ab-a-b > ab-a-b$.
A: Write the class equation for the action of $G$ on $\{1,2,\dots , 999\}$. We get
$999=|\{$fixed points of the action$\}|+\sum [G: stab(n)]$
where the sum is taken over a system of representatives for the non-trivial orbits of the action. The possibilities for $[G: stab(n)]$ in the sum are thus 11 and 101 since $1111>999$, but 999 cannot be written $11a+101b$ with $a,b \geq 0$ as you can see by finding the general solution to $11a+101b=999$. So, we cannot have a non-trivial number of fixed points for the action.
Note: We did not need to assume $G$ was abelian, although from Sylow theory, since 101 is not congruent to 1 modulo 11, it's easy to see that any group of order 1111 is abelian (cyclic even).
