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Let $A$ be a given nonempty set. $S(A)$ is a group with respect to mapping composition. For a fixed element $a$ in $A$, let $H_{a}$ denote the set of all $f \in S(A)$ such that $f(a) = a$. Prove that $H_{a}$ is a subgroup of $S(A)$.

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I don't suppose you have any thoughts of your own? – Zev Chonoles Apr 8 '12 at 23:56
You haven't said what $S(A)$ is. What have you tried? Do you know how to check whether something is a subgroup? For example, you have to check that a product of two elements of $H_a$ is in $H_a$. What does that mean? – Dylan Moreland Apr 8 '12 at 23:57
Sorry. Let $S(A)$ denote the set of all permutations on $A$ – user28615 Apr 9 '12 at 0:02

This sounds like homework, so only a hint. If $f,g \in H_a$, what is $f(g(a))$? Similarly, what is $f^{-1}(a)$? What does this tell you about $f \circ g$ and $f^{-1}$?

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