Let $A = \{ 2, 5, 9\}$. Let $K = \{ F ∈ S_9 | F(a) = a, ∀a ∈ A \}$. Determine order of $K$.

Question

Question: See title

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First off, what is K exactly? My guess: it's just the identity function with order $1$. If it's not the identity function, how would I calculate the order?

Thanks in advance.

Answer 1

Hint: An arbitrary element $\alpha \in F$ has the form: $$\alpha = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ x_1 & 2 & x_3 & x_4 & 5 & x_6 & x_7 & x_8 & 9 \end{pmatrix} $$

where $x_1, x_3, x_4, x_6, x_7, x_8$ is some permutation of the numbers in the set $\{1, 3, 4, 6, 7, 8\}$.

Answer 2

It is all about cosets of $Sym(9)$. Lets look at easier example:

Take the set A = $\{f\in Sym(9)$ : $f(2)=3 \}$.

Actually A=$(23)Sym\{ 1,3,4,5,6,7,8,9\}=(23) \langle (13),(13456789)\rangle$. Every set which is look like A is actually a coset.

Look at the set B= $\{f\in Sym(9)$ : $f(2)=2 \}$. B is exactly $(22)Sym\{1,3,4,5,6,7,8,9\} = \langle (13),(13456789)\rangle$.

For your question you look

$$\{f\in Sym(9) : f(2)=2 \} \cap \{f\in Sym(9) : f(3)=3 \} \cap \{f\in Sym(9) : f(9)=9 \} = (22)(33)(99) Sym\{1,4,5,6,7,8\}$$ which has 6 coset. You can find them by multiply $Sym\{1,4,5,6,7,8\}$ with $ Id,(23),(39),(29),(239),(293)$.I think the analogy is clear.

You can also find the set $\{ f\in Sym(9): f[\{1,2,3\}]=\{1,2,3\} \}$.