# No subgroup of $S_n$ containing stabilizier of 1?

Is it true that the stabilizer of $1\in \left\{1,\dots ,n \right\}$ in $S_n$ is a maximal subgroup? Intuitively I'm thinking that as soon as you add another permutation, you'll somehow be able to construct transpositions, but I don't know what to make of this..

• This is at least true when $n$ is prime. The stabilizer of $1$, $\operatorname{Stab}(1)$, is isomorphic to $S_{n-1}$. If there exists a subgroup $H$ with $\operatorname{Stab}(1) \subsetneq H$, then we must have $|H| > (n-1)!$ and $|H|$ a proper divisor of $|n!|.$ When $n$ is prime, this is impossible. – Kaj Hansen Jul 20 '16 at 16:17
• @KajHansen actually, in my case, $n$ necessarily is not prime :\ – user153312 Jul 20 '16 at 16:18
• Yep, I figured. – Kaj Hansen Jul 20 '16 at 16:18
• – lhf Jul 20 '16 at 16:20

The orbit-stabilizer theorem should work.

Let $G = S_n$, and let $H = \operatorname{Stab}_G(1)$ be the subgroup of permutations fixing $1$, so that $H \cong S_{n - 1}$. Let $\sigma \in G$ be any permutation that doesn't fix $1$.

We consider the group $H' = \langle H, \sigma \rangle \leq G$.

Certainly $H'$ acts on $[n] = \{1, 2, \ldots, n \}$, and under this action, $\operatorname{Orb}_{H'}(1) = [n]$. This is because the permutation $\sigma$ sends $1$ to $\sigma(1) \in \{2,3, \ldots, n\}$, while $H$ acts transitively on this set, so we can send $1$ anywhere (for example: If $\sigma(1) = 3$, then the composition $(3\ 5) \circ \sigma \in H'$ sends $1$ to $5$, so $5$ is in the orbit, and so on).

The stabilizer $\operatorname{Stab}_{H'}(1) = H$, since by definition, every permutation in $G$ fixing $1$ is in $H$; we don't get any new $1$-fixing permutations from $H' = \langle H, \sigma \rangle$.

Thus, $|H'| = n \cdot (n - 1)! = n!$, and so $H'$ must be the full symmetric group on $n$ letters.