\chapter{Lecture 6 May 14th 2018}
\label{chp:lecture_6_may_14th_2018}
% chapter Lecture 6 May 14th 2018
\section{Subgroups (Continued 2)}
\label{sec:subgroups_continued_2}
% section Subgroups (Continued 2)
\subsection{Alternating Groups}
\label{sub:alternating_groups}
% subsection Alternating Groups
Recall that $\forall \sigma \in S_n$, with $\sigma \neq \epsilon$, $\sigma$ can be uniquely decomposed (up to the order) as disjoint cycles of length at least $2$. We will now present a related concept.
\begin{defn}[Transposition]\label{defn:transposition}
\index{Transposition}
A \hlnoteb{transposition} $\sigma \in S_n$ is a cycle of length $2$, i.e. $\sigma = \begin{pmatrix} a & b \end{pmatrix}$, where $a, b \in \{1, ..., n\}$ and $a\ neq b$.
\end{defn}
\begin{eg}
We have that\sidenote{If we apply the permutations on the right hand side, we have that
\begin{gather*}
1 \quad 2 \quad 3 \quad 4 \quad 5 \\
\downarrow \\
1 \quad 2 \quad 3 \quad 5 \quad 4 \\
\downarrow \\
1 \quad 4 \quad 3 \quad 5 \quad 2 \\
\downarrow \\
2 \quad 4 \quad 3 \quad 5 \quad 1
\end{gather*}
}
\begin{equation*}
\begin{pmatrix} 1 & 2 & 4 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 4 & 5 \end{pmatrix}
\end{equation*}
Also, we can show that\sidenote{
\begin{ex}
Show that \autoref{eq:transposition_eg} is true.
\end{ex}
\begin{ex}
Play around with the same idea and create a few of your own transpositions. Note that you will only be able to get an odd number of tranpositions (why?).
\end{ex}
}
\begin{equation}\label{eq:transposition_eg}
\begin{pmatrix} 1 & 2 & 4 & 5 \end{pmatrix} = \begin{pmatrix} 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 5 \end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & 4 \end{pmatrix}
\end{equation}
\end{eg}
Observe that the factorization into transpositions are \hlimpo{not unique or disjoint}. However, the following property is true.
\begin{thm}[Parity Theorem]\label{thm:parity_theorem}
\index{Parity Theorem}
If a permutations $\sigma$ has $2$ factorizations
\begin{equation*}
\sigma = \gamma_1 \gamma_2 \hdots \gamma_r = \mu_1 \mu_2 \hdots \mu_s,
\end{equation*}
where each $\gamma_i$ and $\mu_j$ are transpositions, then $r \equiv s \mod 2$.
\end{thm}
\marginnote{
I have referred to the following source: \url{https://www.maa.org/sites/default/files/images/upload_library/4/vol1/parity/ParityJOKHistory.html}. In the literature review of the author, there are (at least) 6 approaches to proving the statement, some argued to be better or more intutive than the other. Recent proofs, as mentioned in the article, use a reduction method (or algorithm) to proof an alternate version of our statement. I intended to study the proof, but ran short on time, and so I shall present my understanding of the proof provided in Dummit and Foote's 3rd Edition of Abstract Algebra. My opinion of the proof presented in the said book is that it is not immediately intuitive and relies on subtle connections to the actual statement, which is why I looked into other sources and found the source above.
For the proof provided by Dummit and Foote, as well as in some of the other proofs that I have come across, the Parity Theorem is presented as a statement that is different from, but equivalent to, our statement here.
}
\begin{proof}
Let $x_1, x_2, ..., x_n$ be distinct variables. Let $\Delta$ be the following product:
\begin{equation*}
\Delta = \prod_{1 \leq i < j \leq n} (x_i - x_j).
\end{equation*}
For each $\sigma \in S_n$, let $\sigma$ act on $\Delta$ by permuting the indices of the variables so as to permute the variables themselves, i.e.
\begin{equation*}
\sigma( \Delta ) = \prod_{1 \leq i < j \leq n} ( x_{\sigma(i)} - x_{\sigma(j)} ).
\end{equation*}
Then for each $(x_i - x_j)$ in the product $\Delta$, after applying $\sigma$, we have that the result is either $(x_k - x_l)$ or $(x_l - x_k)$ for $1 \leq k < l \leq n$ but not both. In $\sigma(\Delta)$, for each of the factors, if we have $(x_l - x_k)$ for $1 \leq k < l \leq n$, rewrite that factor as $- (x_k - x_l)$. Now if we collect all the $(-1)$'s, we get that $\sigma(\Delta) = \pm \Delta$ depending on \textbf{whether if there is an odd or even number of factors that are of the form $(x_l - x_k)$ with $k < l$}. From here, we can use the definition of a sign of a permutation (as introduced in class on May 30th, 2018) and write the sign as
\begin{equation*}
\sign(\sigma) = \begin{cases}
1 & \text{if } \sigma(\Delta) = \Delta \\
-1 & \text{if } \sigma(\Delta) = -\Delta
\end{cases}.
\end{equation*}
As mentioned in class, the sign of a permutation is a homomorphism. Now for each $\sigma \in S_n$, we can express the permutation as a product of disjoint cycles. Consider the simplest case where $\sigma$ is a permutation with only one cycle. We know that we can rewrite $\sigma$ as a product of transpositions. Suppose $\sigma = \gamma_1 \gamma_2 ... \gamma_r$ for some $r > 0$, where each $\gamma_i$, $1 \leq i \leq r$, is a transposition. We then have that
\begin{equation*}
\sign(\sigma) = \sign(\gamma_1) \sign(\gamma_2) ... \sign(\gamma_r)
\end{equation*}
Note that since transpositions are odd permutations, we essentially have
\begin{equation*}
\sign(\sigma) = (-1)^r.
\end{equation*}
We have that if $\sign(\sigma) = 1$, then $r$ must be even, which coincides with our bolded argument above. Similarly, if $\sign(\sigma) = -1$, we have that $r$ must be odd. Therefore, if we have that
\begin{equation*}
\sigma = \gamma_1 \gamma_2 ... \gamma_r = \mu_1 \mu_2 ... \mu_s,
\end{equation*}
where $s > 0$ and the $\mu_j$'s, for $1 \leq j \leq s$, are transpositions, $r$ and $m$ must either be both even or both odd. In other words, $r \equiv s \mod 2$.
For cases with more than one cycle, we can consider the individual cycles and the homomorphicity of the sign will extend our above argument for permutations that is a product of more than one disjoint cycle.\qed
\end{proof}
\begin{defn}[Odd and Even Permutations]\label{defn:odd_and_even_permutations}
\index{Odd Permutations}\index{Even Permutations}
A permutation $\sigma$ is even (or odd) if it can be written as a product of an even (or odd) number of transpositions. By \autoref{thm:parity_theorem}, a permutation must either be even or odd, but not both.
\end{defn}
\begin{thm}[Alternating Group]\label{thm:alternating_group}
\index{Alternating Group}
For $n \geq 2$, let $A_n$ denote the set of all even permutations in $S_n$. Then
\begin{enumerate}
\item $\epsilon \in A_n$
\item $\forall \sigma, \tau \in A_n \enspace \sigma \tau \in A_n$ and $\exists \sigma^{-1} \in A_n$ such that $\sigma \sigma^{-1} = \epsilon = \sigma^{-1} \sigma$
\item $\abs{A_n} = \frac{1}{2} n!$
\end{enumerate}
\end{thm}
\begin{note}
From items 1 and 2, we know that $A_n$ is a subgroup of $S_n$. $A_n$ is called the \hlnoteb{alternating subgroup of degree $n$}.
\end{note}
\begin{proof}
\begin{enumerate}
\item We have that $\epsilon = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}$. Thus $\epsilon$ is even and so $\epsilon \in A_n$.
\item $\forall \sigma, \tau \in A_n$, we may write
\begin{align*}
\sigma &= \sigma_1 \sigma_2 \hdots \sigma_r \quad \text{and} \\
\tau &= \tau_1 \tau_2 \hdots \tau_s,
\end{align*}
where $\sigma_i, \tau_j$ are transpositions, and $r, s$ are even integers. Then
\begin{equation*}
\sigma \tau = \sigma_1 \sigma_2 \hdots \sigma_r \tau_1 \tau_2 \hdots \tau_s
\end{equation*}
is a product of $(r + s)$ transpositions, and thus $\sigma \tau$ is even. Thus $\sigma \tau \in A_n$.
For the inverse, note that since $\sigma_i$ is a transposition, we have that $\sigma_i^2 = \epsilon$ and thus $\sigma_i^{-1} = \sigma_i$. It follows that
\begin{align*}
\sigma^{-1} &= (\sigma_1 \sigma_2 \hdots \sigma_r)^{-1} \\
&= \sigma_r^{-1} \sigma_{r - 1}^{-1} \hdots \sigma_2^{-1} \sigma_1^{-1} \\
&= \sigma_r \sigma_{r - 1} \hdots \sigma_2 \sigma_1
\end{align*}
which is an even permutation and
\begin{equation*}
\sigma \sigma^{-1} = \sigma_1 \sigma_2 \hdots \sigma_r \sigma_r \hdots \sigma_2 \sigma_1 = \epsilon.
\end{equation*}
Thus $\exists \sigma^{-1} \in A_n$ such that it is the inverse of $\sigma$.
\item Let $O_n$ denote the set of odd permutations in $S_n$.\marginnote{For the proof of 3, we know that $\abs{S_n} = n!$, which is twice of the suggested order of $A_n$. Since we took out the even permutations of $S_n$, we just need to make the rest of the permutations, the odd permutations, into a set and prove that $A_n$ and this new set has the same size. One way to show this is by creating a bijection between the two.
Also, note that the set of all odd permutations of $S_n$ is not a group, since
\begin{itemize}
\item there is no identity element in this set; and
\item this set is not closed under map composition.
\end{itemize}
We have shown that $\epsilon$ is an even permutation, and so by the \hyperref[thm:parity_theorem]{Parity Theorem}, it cannot be an odd permutation, and there is only one identity in $S_n$. The set is not closed under map composition since if we compose two odd permutations, we would get an even permutation, which does not belong to this set.
} Then we have $S_n = A_n \cup O_n$, and by the \hyperref[thm:parity_theorem]{Parity Theorem}, we have that $A_n \cap O_n = \emptyset$. Since $\abs{S_n} = n!$, to prove that $\abs{A_n} = \frac{1}{2} n!$, it suffices to show that $\abs{A_n} = \abs{O_n}$.
Let $\gamma = \begin{pmatrix} 1 & 2 \end{pmatrix}$ and $f : A_n \to O_n$ such that $f(\sigma) = \gamma \sigma$. Since $\sigma$ is even, $\gamma \sigma$ is odd, and so $f$ is well-defined.
Also, if $\gamma \sigma_1 = \gamma \sigma_2$, then by \hyperref[propo:cancellation_laws]{Cancellation Laws}, $\sigma_1 = \sigma_2$, and hence $f$ is injective.
Finally, $\forall \tau \in O_n$, we have that $\gamma \tau = \sigma \in A_n$. Note that
\begin{equation*}
f(\sigma) = \gamma \sigma = \gamma \gamma \tau = \tau.
\end{equation*}
Therefore, $f$ is surjective.
It follows that $\abs{A_n} = \abs{O_n}$. \qed
\end{enumerate}
\end{proof}
% subsection Alternating Groups (end)
\subsection{Order of Elements}
\label{sub:order_of_elements}
% subsection Order of Elements
\begin{notation}
If $G$ is a group and $g \in G$, we denote
\begin{equation*}
\lra{g} = \{ g^k : k \in \mathbb{Z} \}.
\end{equation*}
Note that $1 = g^0 \in \lra{g}$.
If $x = g^m, y = g^n \in \lra{g}$ where $m, n \in \mathbb{Z}$, then
\begin{equation*}
xy = g^m g^n = g^{m + n} \in \lra{g}
\end{equation*}
and we have $\exists x^{-1} = g^{-m} \in \lra{g}$ such that
\begin{equation*}
xx^{-1} = g^m g^{-m} = g^0 = 1.
\end{equation*}
\end{notation}
Along with the \hlnoteb{Subgroup Test}, we have the following proposition:
\begin{propo}[Cyclic Group as A Subgroup]\label{propo:cyclic_group_as_a_subgroup}
If $G$ is a group and $g \in G$, then $\lra{g}$ is a subgroup of $G$.
\end{propo}
\begin{defn}[Cyclic Groups]\label{defn:cyclic_groups}
\index{Cyclic Group}
Let $G$ be a group and $g \in G$. Then we call $\lra{g}$ the \hlnoteb{cyclic subgroup} of $G$ generated by $g$. If $G = \lra{g}$ for some $g \in G$, then we say that $G$ is a \hlnoteb{cyclic group}, and $g$ is a \hldefn{generator} of $G$.
\end{defn}
% subsection Order of Elements (end)
% section Subgroups (Continued 2) (end)
% chapter Lecture 6 May 14th 2018 (end)