# Abstract Nonsense

## Halmos Section 28: Parity

Point of post: In this post I will solve the, very few, problems in the twenty-eighth section of Halmos. This section is on the parity (sign) of a permutation.

1.

Problem: How many elements of $A_n$ are there? (here $A_n$ denotes the Alternating Group)

Proof: Let $O_n$ be the set of all odd permutations in $S_n$ and define

$f:O_n\to A_n:\pi\mapsto (1,n)\pi$

We claim that this mapping is bijective. It’s clearly injective since $f(\pi)=f(\pi')\implies (1,n)\pi=(1,n)\pi'\implies \pi=\pi'$. To see it’s surjective we note that if $\sigma\in A_n$ then $\text{sgn}(\sigma)=1$ and thus

$\text{sgn}\left((1,n)\sigma)\right)=\text{sgn}\left((1,n)\right)\text{sgn}\left((1,n)\right)=-1\cdot 1=-1$

so that $(1,n)\sigma\in O_n$. But, then $f\left((1,n)\sigma\right)=(1,n)^2\sigma=\sigma$. It follows that $f$ is bijective. Note though that since $S_n=O_n\sqcup A_n$ ($\{A_n,O_n\}$ partition $S_n$) we may conclude that

$n!=\left|S_n\right|=\left|O_n\cup A_n\right|=\left|O_n\right|+\left|A_n\right|=2\left|O_n\right|=2\left|A_n\right|$

and thus $\displaystyle \left|A_n\right|=\left|O_n\right|=\frac{n!}{2}$.

2.

Problem: Give examples of even permutations with even order and even permutations with odd order; do the same for odd permutations.

Proof: First note that $(1,2,3)\in S_3$ is an even permutation (as can be checked, I did it by computing $P_{(1,2,3)}$ and finding that its determinant is $1$) yet $\text{ord}\left((1,2,3)\right)=3$ (or, now that I think about it, the identity map $\text{id}_{[n]}$ is even since $\text{id}_{[n]}=(1,2)(1,2)$ but its order is $1$). Take $(1,2)(3,4)\in S_4$, this is even and has even order.

Note that $(1,2)\in S_3$ is odd but has even order.  Take, $(1,2,3)(4,5,6)(7,8,9)\in S_9$. This is odd and has odd order.

3.

Problem: Prove that $(1,2,3),\cdots,(1,2,n)$ generates $A_n,\text{ }n\geqslant 3$.

Proof: We proceed by induction. Clearly this i true for $n=3$. Now, suppose then that it’s true for $n=m$. Then, note that if $\pi\in A_{m+1}$ with $\pi(m+1)=k$,  then $\pi(1,2,k)(1,2,m+1)^{-1}$ is even and leaves $m+1$ fixed and thus by our induction hypothesis may be written as the product of $3$-cycles, and thus $\pi$ may be written as the product of three cycles. The conclusion follows.

Remark: Note that in general that if $(1,\cdots,n)\in S_n$ that $\text{sgn}\left((1,\cdots,n)\right)=(-1)^{n-1}$. To see this note that a quick computations shows that



from where it follows that $P_{(1,\cdots,n)}$ can be written as as the product of $n-1$ row-changing matrices and thus $\det P_{(1,\cdots,n)}=\text{sgn}\left((1,\cdots,n)\right)=(-1)^{n-1}$

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print