MHB Symmetrical Group: Prove Properties & Find Element Count

[qamaz]

$$\text{ Let } n∈ \mathbb{N} \text{ and } S_{n} \text{ symmetrical group on } \underline n\underline .

\text{ Let }
π ∈ S_{n} \text{ and z } \text{ the number of disjunctive Cycles of π. Here will be counted 1 - Cycle }. (a) \text{ Prove that } sgn (π) = (-1)^{n-z}.

(b) \text{ Prove that subset } A_{n}= \{π∈Sn∣sgn(π)=1\} ⊆ S_{n}\text{ is subgroup of } S_{n}.

(c)

\text{ Find number of elements } |A_{n}| \text{ of a subgroup } A_{n} \text{ from (b) }

$$

 

[Olinguito]

I’ll just do part (a); the rest I leave to you. Note a few things to start with:

• Even cycles (cycles of even length) are odd permutations whereas odd cycles (cycles of odd length) are even permutations.
  —
• Let $l_e$ be the sum of the lengths of the even cycles, and similarly $l_o$ for the odd cycles; let $z_e$ and $z_o$ be the number of even and of odd cycles respectively. Thus $n=l_e+l_o$ and $z=z_e+z_o$. Then note that $l_e$ is always even; on the other hand $l_o$ has the same sign as $z_o$. (If there is an even number of odd cycles, the sum of their lengths is even; if there is an odd number of odd cycles, the sum of their lengths is odd.) Also, the fact that $l_e$ is always even and $n=l_e+l_o$ implies that $n$ and $l_o$ have the same parity (both odd or both even).
  —
• $\mathrm{sgn}(\pi)$ depends only on $z_e$ and $\mathrm{sgn}(\pi)=\mathrm{sgn}(z_e)$. In other words, $\pi$ is an even permutation if there is an even number of even cycles and it’s an odd permutation if there is an odd number of even cycles. (Make sense? See observation (i) above.)

First, suppose that $n$ is even. Then $l_o$ is even and so $z_o$ is even (observation (ii) above). Hence
$$\mathrm{sgn}(n-z)=\mathrm{sgn}(z)=\mathrm{sgn}(z_e+z_o)=\mathrm{sgn}(z_e)=\mathrm{sgn}(\pi)$$
by observation (iii) above.

Now suppose $n$ is odd. Then $l_o$ is odd and so $z_o$ is odd (observation (ii)) and so $l_o-z_o$ is even. Then
$$\begin{array}{rcl}\mathrm{sgn}(n-z) &=& \mathrm{sgn}([l_e+l_o]-[z_e+z_o]) \\ {} &=& \mathrm{sgn}(z_e)\ \text{(since }l_e\ \text{and}\ l_o-z_o\ \text{are both even)} \\ {} &=& \mathrm{sgn}(\pi)\end{array}$$
again. QED.