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Let $S$ be any set, $A(S)$ the set of one-to-one mappings of $S$ onto itself, made into a group under the composition of mappings. If $x_0 \in S$, what is meant by $H(x_0) = \left\{\phi \in A(S): x_0 \phi = x_0\right\}$? The set that contains the element $\phi$ in $A(S)$ that maps $x_0$ onto itself?
How does one verify that something like this is a subgroup of $A(S)$?
For any $\phi, \phi' \in H(x_0)$, we have $x_0\phi \phi' = x_0 \phi' = x_0 $. Thus $H(x_0)$ is closed under multiplication. How do I show that for any $\phi \in H(x_0)$ we have $\phi^{-1} \in H(x_0)$?
Also if for $x_1 \ne x_0 \in H(x_1)$, we similarly define $H(x_1)$, what's $H(x_0) \cap H(x_1)$?
How does one verify that something like this is a subgroup of $A(S)$?
For any $\phi, \phi' \in H(x_0)$, we have $x_0\phi \phi' = x_0 \phi' = x_0 $. Thus $H(x_0)$ is closed under multiplication. How do I show that for any $\phi \in H(x_0)$ we have $\phi^{-1} \in H(x_0)$?
Also if for $x_1 \ne x_0 \in H(x_1)$, we similarly define $H(x_1)$, what's $H(x_0) \cap H(x_1)$?
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