Proving that a set has a fixed point

Question

Let $G$ be a group of order $16$, and $\pi$ a $G$ action on a set $X$, where $|X|=5$.

Prove that $X$ has a fixed point.

I can see that $|X|\not\equiv0$ mod $2$ and I should assume that there is no fixed point and get a contradiction - $|X|\equiv 0$ mod $2$, but now sure how...

Answer 1

Size of orbits divide the order of the group (comes from Orbit-Stabilizer Lemma). So, your orbits should be of size 1, 2, 4, 8, or 16.

The orbit sizes must add up to $|X|$ since the orbits partition the set. Try to add up to 5 using 1, 2, 4, 8, 16. You get only these possibilities: 1+1+1+1+1, 1+1+1+2, 1+2+2, 1+4

So one of the orbits must be size 1, which means it's a fixed point.

See A question about the fixed point and group action.