In Lang's Algebra, the following two exercises are presented to the reader in the following order:
Groups Exercise 15: Let $G$ be a finite group acting on $S$, a finite set of at least $2$ elements. Assume there is only one orbit. Prove there is at least one $g\in G$ with no fixed point, i.e. for all $s\in S$, $g\cdot s\ne s$.
Groups Exercise 19b.: Let $G$ be a finite group action on a finite set $S$. For each $g\in G$, define ${\rm Stab}(g):=\{s\in S : gs = s\}$. Prove the number of orbits of $S$ is equal to $$ \frac{1}{\#G}\sum_{g\in G}\#{\rm Stab}(g). $$
I can include the proofs I have for both of these if desired to assuage any concerns that I am fishing for the community to give me free proof(s).
My question is as follows: The latter exercise appears to just be Burnside's Lemma, which I am all for having as an exercise. But the former exercise to me screams "Hey, this is the type of problem where Burnside's Lemma does a lot of the heavy lifting." Am I missing so obvious (or clever) approach to the former that allows one to circumvent using Burnside's Lemma, or should I chalk this up to the order of exercises not really mattering in the grand scheme of pedagogy?
Thank you all for your help and insight :D
I agree that Burnside's Lemma is a great way to solve Exercise 15, but there are other ways to solve it that do not use Burnside's Lemma in an obvious way. Let me describe one such way.
Let $S=\{s_1,\ldots,s_n\}$ and let $H=\textrm{Stab}(s_1)$. Note that $H\lneq G$ is a proper subgroup of $G$, since $[G:H]=|Gs_1|=|S|\geq 2$. The set of elements of $G$ that fix some element of $S$ is $\bigcup_{i=1}^n \textrm{Stab}(s_i)$. The set of elements that fix some element of the orbit $Gs_1$ is $\bigcup_{g\in G} gHg^{-1}$. Since we are assuming that $S$ is a single orbit we have $\bigcup_{i=1}^n \textrm{Stab}(s_i)=\bigcup_{g\in G} gHg^{-1}$. The union of conjugates of a proper subgroup $H\lneq G$ is a proper subset of $G$, so there must exist some $g\in G-\bigcup_{g\in G} gHg^{-1}$ = $G-\bigcup_{i=1}^n \textrm{Stab}(s_i)$, and such a $g$ fixes no element of $S$.