Subgroups of $S_n$ with exactly one fixed point for each element all have the same fixed point. Let $G$ be a subgroup of $S_n$ (where $n$ is a positive integer) such that each non identity element $g\in G$ has exactly one fixed point. Prove there is an element of $[n]$ that is fixed by every permutation of $G$.
What I am trying to do is prove it by contrapositive, showing that if for every $k\in [n]$ there is a permutation that moves $k$ then there is a permutation with no fixed points, but it is not clear to me how I can do this.
 A: Apply Burnside's Lemma:
Consider the action of $G$ over $[n]$. The number of orbits is
$$|[n]/G|=\sum_{g\in G}\frac{[n]^g}{|G|}$$
where $[n]^g=\{k\in[n]:g(k)=k\}$.
The hypothesis says that $|[n]^g|=1$ for all $g\neq e$. Moreover, $[n]^e=[n]$. Then
$$|[n]/G|=\frac n{|G|}+\sum_{g\in G,g\neq e}\frac{1}{|G|}=\frac{|G|+n-1}{|G|}>1$$
This means two things:


*

*There are at least two orbits.

*$|G|$ divides $n-1$, say $k|G|+1=n$. Note that the number of orbits is $k+1$.


Now, $n$ is the sum of the cardinals of the orbits. Moreover, there is a stabilizer subgroup $I_x$ associated to each orbit $O_x$, and $|G|=|I_x|\cdot|O_x|$. This yields
$$n=\sum\frac{|G|}{|I_x|}$$
where the sum spans the set of orbits and hence, it has $k+1$ terms. It also can be written:
$$\frac n{(k+1)|G|}=\frac1{k+1}\sum\frac1{|I_x|}$$
We see that LHS is near from $1$, so there must be many terms in the sum at RHS that are $1$. Let's see how many.
For that, let's apply the following:

Lemma. If $a_1,\ldots,a_s$ are positive integers and exactly $r$ of them are $1$, then $$\frac1s\sum_{j=1}^s\frac1{a_j}\le\frac{s+r}{2s}$$
Proof: $$\frac1s\sum_{j=1}^s\frac1{a_j}\le\frac rs+\sum_{j=1}^{s-r}\frac1{2s}=\frac{s+r}{2s}$$

So let $r$ be the number of orbits of size $|G|$ (that is, the number of stabilizers of size $1$). Then
$$\frac n{(k+1)|G|}\le \frac{k+1+r}{2k+2}$$
Solving for $r$ gives
$$r\ge\frac{2n}{|G|}-(k+1)=k-1+\frac2{|G|}$$
which leads to
$$r\ge k$$
Since $r$ is the number of orbits of size $|G|$, $r\le k+1$, but $r\neq k+1$ since $(k+1)|G|>k|G|+1=n$, so $r=k$.
To sum up, there are $k$ orbits of size $|G|$. The size of the other orbit is
$$n-k|G|=1$$
which completes the proof.
A: Your claim is not true. For example, consider the subgroup $G < S_5$ generated by $(123)$ and $(12)(45)$, which is isomorphic to $S_3$. Here every element of $G$ fixes a point but there is no common fixed point for $G$.
