Subgroups of $S_n$ with exactly one fixed point for each element all have the same fixed point.

Solution 1:

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.

Solution 2:

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 $G$.