**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.15**

Fix $i \in \{ 1, \ldots, n \} = A$, and let $S_n$ act on $A$ in the natural way. Show that $\mathsf{stab}_{S_n}(i) \cong S_{n-1}$.

Solution: It is clear that $$\mathsf{stab}_{S_n}(i) = \{ \sigma \in S_n \ |\ \sigma(i) = i \}.$$ Removing from each function in this set the pair $(i,i)$, we have the full symmetric group on $\{ 1, 2, \ldots, i-1,i+1,\ldots,n\}$. We saw in a previous exercise that the isomorphism type of $S_A$ depends only on the cardinality of $A$, so that the conclusion holds.