If $G$ acts on $\Omega$ and $|G| = |\Omega| + 1$, show there exists nontrivial element fixing point without Burnside's lemma Let $G$ act on $\Omega$ transitively, and let $|G| = |\Omega| + 1$ (both sets are assumed to be finite). I want to show from first principles (using maybe arguments like the pigeonhole principle, but not Burnside's lemma) that there exists a non-trivial element having a fixed point. For example let $\Omega = \{\alpha_1, \ldots, \alpha_n\}$ and $G = \{1,g_1,\ldots, g_n\}$, write w.l.o.g.
$\alpha_1^{g_1} = \alpha_2,\alpha_1^{g_2} = \alpha_3,\ldots, \alpha_1^{g_{n-1}} = \alpha_n$, then
$$
 \alpha_1^{g_n} \in \{\alpha_1,\ldots, \alpha_n\}
$$
so if $\alpha_1^{g_n} = \alpha_1$ we have found a fixed point for $g_n$, otherwise $\alpha_1^{g_{n-1}} = \alpha_i = \alpha_1^{g_{i-1}}$ for some $i = 2,\ldots, n-1$, hence $\alpha_1^{g_{n-1} g_{i-1}^{-1}} = \alpha_1$, but I still have to exclude that $g_{n-1} g_{i-1}^{-1} = 1$, i.e. the elements are inverse to each other. Maybe finding an enumeration scheme like the one above such that this case is excluded. So any ideas, could we show $g_{n-1} g_{i-1}^{-1} \ne 1$?
Remark: Of course by Burnside's lemma we would have $|\Omega|+1 = |G| = \sum_{x\in G}\mbox{fix}(x)$, hence with $\mbox{fix}(1) = \Omega$ there has to exist exactly one element $x \ne 1$ with $\mbox{fix}(x) = \{\alpha\}$.
 A: Note that $g_{n-1}\neq g_{i-1}$ so there is some element $\omega\in\Omega$ for which $g_{i-1}(\omega)\neq g_{n-1}(\omega)$ and so $g_{i-1}^{-1}g_{n-1}(\omega)\neq\omega$ whence $g_{i-1}^{-1}g_{n-1}$ cannot be the identity. 
A: $G$ acts transitively on $\Omega$ (both finite). If the action were free then $|G| = |\Omega|$. Thus by your assumption the action is not free, i.e. for some $\omega \in \Omega$ the set $Stab(\omega) = \{g \in G \mid g \cdot \omega = \omega \}$ contains at least two elements, thus some non-trivial element $g \in G$ fixing $\omega$.
A: If a finite group $G$ acts transitively on $\Omega$ then $|\Omega|$ is a divisor of $|G|$ by the orbit-stabilizer theorem, and so if $|\Omega|=|G|-1$ then we know $(|G|-1)\mid |G|$ which implies $|G|=2$ and $|\Omega|=1$, and hence the only element of $\Omega$ must be a fixed point.
In any case, pick $\omega\in\Omega$ and consider the map $G\to G\omega$. Since the range is smaller than the domain, by the pigeonhole principle there must be a collision, i.e. $g\omega=h\omega$ for a pair of distinct group elements $g$ and $h$, hence every point $\omega$ is fixed by some group element $g^{-1}h$. (There is no reason to choose an ordering on the group elements.)
