Stabilizer in Group of Transformations

Theorem

Let $X$ be any set with $n$ elements (where $n \in \Z_{>0}$).

Consider the symmetric group on $n$ letters $S_n$ as a group of transformations on $X$.

Let $x \in X$.

Then the stabilizer of $x$ is isomorphic to $S_{n - 1}$.

Proof

Consider the initial segment of natural numbers $\N_n = \set {1, 2, \ldots, n}$.

By the definition of cardinality, $H$ is equivalent to $\N_n$.

Without loss of generality we can consider $S_n$ acting directly on $\N_n$.

The stabilizer of $n$ in $\N_n$ is all the permutations of $S_n$ which fix $n$, which is clearly $S_{n - 1}$.

A permutation can be applied to $\N_n$ so that $i \to n$ for any $i$.

Thus one can build an isomorphism to show the result for a general $i$.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.6$. Stabilizers: Example $107$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$