# Every Transitive Permutation Group Has a Fixed Point Free Element

If $G$ acts transitively by permutations on a finite set $A$ with more than one element (i.e. $G$ is a transitive permutation subgroup of the symmetric group $S_A$). Why does $G$ necessarily contain an element which has no fixed points (i.e. $g$ such that $g \cdot a \neq a$ for any $a \in A$)?

The hint I have is to think about, given $a \in A$, what fraction of elements of $G$ fixes $a$. I'm not sure how to go about this hint...

By Burnside's lemma, you have

$$\frac{1}{|G|}\sum_{g \in G}|\text{fix }g| = |A/G|=1$$

Since $1 \in G$ has $|A|>1$ fixed points, at least one of the terms in the sum must be $0$.

Here's another solution, in case you haven't heard about Burnside's lemma:

Given $a \in A$, the subset $G_a \subseteq G$ which fixes $a$ (called the stabilizer of $a$ in $G$) forms a subgroup of $G$. Since $G$ acts transitively, each of these point stabilizers is conjugate, i.e. for each $b \in A$ we have $G_b = g^{-1}G_ag$ for some $g \in G$. If we suppose that every element of $G$ has a fixed point, then each $\sigma \in G$ is contained in some stabilizer $G_b$. Then $\sigma \in \bigcup_{g\in G} g^{-1}G_ag$ for all $\sigma \in G$. But this can't happen since $G$ cannot be the union of the conjugates of any proper subgroup. (See if you can prove this.)

• Actually, Bruno's argument is a very nice proof of the fact that the conjugates of a proper subgroup do not cover the group :) Commented Feb 6, 2012 at 2:12
• Thank you! I got it all! And I learned so many additional things with this exercise (such as that a finite $G$ is never the union of the conjugates of a proper subgroup -- that's interesting!).
– Rick
Commented Feb 6, 2012 at 15:49

It suffices to show that $$\bigcup_{a \in A} \operatorname{Stab}_a \neq G$$ where $$G$$ is the transitive permutation group on $$A$$, so there is some $$\sigma \in G$$ not in $$\bigcup_{a \in A} \operatorname{Stab}_a$$, meaning $$\sigma(a) \neq a$$ for all $$a \in A$$.

As $$G$$ is transitive, the Orbit-Stabilizer theorem says that $$|\operatorname{Stab}_a| = \frac{|G|}{|A|}$$ for all $$a$$. Also every stabilizer contains $$1_G$$, so $$\left| \bigcup_{a \in A} \operatorname{Stab}_a \right| \le 1 + |A|\left(\frac{|G|}{|A|} - 1\right) < |G|,$$ which is what we wanted.

• Just pointing out that the reason the weak inequality holds is because $1_G$ is in every stabilizer subgroup as this was not immediately obvious to me (I thought you were bounding the cardinality of the non-disjoint union by a disjoint union, i.e. $|\bigcup_{a \in A} \operatorname{Stab}_a| \leq \sum_{a \in A} |\operatorname{Stab}_a| = \sum_{a \in A} \frac{|G|}{|A|}$, which doesn't work) Commented Dec 30, 2021 at 19:44
• P.S. this made me realize that the correct inequality can be seen as a general principle for any union of subgroups: for subsets we have the bound $|\bigcup_i X_i|\leq|\bigsqcup_i X_i|=\sum_i |X_i|$ whereas for subgroups we have $|\bigcup_i G_i|\leq|\bigsqcup_i G_i| = 1 + \sum_i(|G_i| - 1)$ where we define subgroups to be disjoint if they intersect only in the identity element, in which case we may use the disjoint union symbol $G_i \sqcup G_j$ (however this union may not be a subgroup, so one will often need to consider $\langle G_i \sqcup G_j \rangle$) Commented Dec 30, 2021 at 19:50
• Thanks @JoelBrennan, I agree I should've mentioned that :D Commented Dec 31, 2021 at 10:33