Return to Homework 7, Glossary, Theorems
Problem 12
(+) Let $G$ be a group of permutations on a set $A$. Let $a\in A$ and define $\textrm{stab}(a) = \{\alpha \in G \ | \ \alpha(a)=a\}$. Then $\textrm{stab}(a)$ is called the $\textbf{stabilizer of}$ $a$ $\textbf{in}$ $G$. Prove that $\textrm{stab}(a)$ is a subgroup of $G$.
Solution
(Closure) Let $\sigma,\tau \in \textrm{stab}(a)$. Then $(\sigma\tau)(a)=\sigma(\tau(a))=\sigma(a)=a$, so $\sigma\tau \in \textrm{stab}(a)$. So, $\textrm{stab}(a)$ is closed.
(Identity) Clearly, the identity permutation, (1), is in $\textrm{stab}(a)$, since it fixes all elements of $A$.
(Inverses) Suppose $\sigma \in \textrm{stab}(a)$. Then we know $\sigma(a)=a$. By the definition of inverse function, then, $\sigma^{-1}(a)=a$, and hence, $\sigma^{-1}\in \textrm{stab}(a)$.
Thus, $\textrm{stab}(a)$ is a subgroup of $G$.
