Question

Question 24: Let ${\mathbf{B}}_{\mathbf{n}}$denote the set of odd permutations in ${\mathit{S}}_{\mathbf{N}}$. Define a function $\mathbf{f}\mathbf{:}{\mathbf{A}}_{\mathbf{n}}\to {\mathbf{B}}_{\mathbf{n}}$by $\mathbf{f}\left(\mathbf{\alpha }\right)=\left(\mathbf{12}\right)\mathbf{\alpha }$.

1. Prove that f is injective.
2. Prove that f is surjective. [Hint: If $\mathbf{\beta }\in {\mathbf{B}}_{\mathbf{n}}$, then ${\mathit{A}}_{\mathbf{n}\mathbf{}}\left(12\right)\mathbf{\beta }\mathbf{\in }{\mathbf{A}}_{\mathbf{n}}$ .] So f is bijective. Hence, role="math" localid="1651768382483" $A_{N}AND{B}_N$have the same number of elements.
3. Show that $\left|A_n\right|\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{2}}$. [Hint: Every element ${\mathit{S}}_{\mathbf{n}\mathbf{}}$of is in ${\mathit{A}}_{\mathbf{n}\mathbf{}}\mathit{o}\mathit{r}\mathbf{}{\mathit{B}}_{\mathbf{n}}$(but not both) and $\left|{\mathbf{S}}_{\mathbf{n}}\right|=\mathbf{n}\mathbf{!}$.]

SeeExercise 39(a) and (b) for a generalization of this exercise.

Short answer

a) It is proved that is injective.

Step-by-step solution

Step-by-Step Solution Step 1: Injective function

The function is calledinjective if for every x and y in A, when $f\left(x\right)=f\left(y\right)$, then $x=y$.

Show that f  is injective

a)

Consider that $\alpha ,\beta \in A_n$and that$f\left(\alpha \right)=f\left(\beta \right)$, namely $\left(12\right)\alpha =\left(12\right)\beta$.

Now, left multiply the equation by ${\left(12\right)}^{-1}$as follows:

role="math" localid="1651768743021" $$\begin{gathered} {\left(12\right)}^{-1}\left(12\right)\alpha ={\left(12\right)}^{-1}\left(12\right)\beta \\ \alpha =\beta \end{gathered}$$

Therefore, is finjective

Hence, it is proved that fis injective.