From 82a2d68e4c02a7ae108e55e449004962ef9e99cf Mon Sep 17 00:00:00 2001 From: Robert <9291792+bushshrub@users.noreply.github.com> Date: Mon, 2 Sep 2024 12:26:05 -0400 Subject: [PATCH] add hk theorem, groups of order 2p theorem and orb stab theorem --- .vscode/settings.json | 3 + chapter-4-permutation-groups.tex | 9 ++ ...lassification-of-finite-abelian-groups.tex | 81 +++++++++++++++++ chapter-lagrange-theorem.tex | 86 ++++++++++++++++++- chapter-normal-subgroups-homomorphisms.tex | 29 ++++++- 5 files changed, 204 insertions(+), 4 deletions(-) create mode 100644 .vscode/settings.json create mode 100644 chapter-classification-of-finite-abelian-groups.tex diff --git a/.vscode/settings.json b/.vscode/settings.json new file mode 100644 index 0000000..082b194 --- /dev/null +++ b/.vscode/settings.json @@ -0,0 +1,3 @@ +{ + "makefile.configureOnOpen": false +} \ No newline at end of file diff --git a/chapter-4-permutation-groups.tex b/chapter-4-permutation-groups.tex index 3a96069..ed4da0a 100644 --- a/chapter-4-permutation-groups.tex +++ b/chapter-4-permutation-groups.tex @@ -436,5 +436,14 @@ \section{Problems} Prove Cayley's Theorem. \end{exercise} +\begin{exercise}[Orbits partition a set] +\label{ex:orbits-partition-set} + Let $G$ be a group acting on a set $S$. Define $\sim$ on $S$ by + \[ + x \sim y \iff x \in \operatorname{orb}_G (y). + \] + Show that $\sim$ is an equivalence relation, and that the equivalence class + of $x$ under $\sim$, $[x]_\sim$ is precisely $\operatorname{orb}_G(x)$. +\end{exercise} \end{document} \ No newline at end of file diff --git a/chapter-classification-of-finite-abelian-groups.tex b/chapter-classification-of-finite-abelian-groups.tex new file mode 100644 index 0000000..a2a81c7 --- /dev/null +++ b/chapter-classification-of-finite-abelian-groups.tex @@ -0,0 +1,81 @@ +\documentclass[./main.tex]{subfiles} + +\begin{document} + + +\section{Classification of finite abelian groups} +Of all the groups, the finite abelian groups are relatively nice behaved, +because they're abelian. What is even nicer behaved are the finite cyclic +groups. Since they're so nicely behaved, it would be nice if we could understand +everything about them. The theorem we present in this chapter will go a long way +to dealing with this. + +To motivate the theorem, recall that the fundamental theorem of arithmetic tells +us that every number can be factorized uniquely as a product of primes, i.e. $n += p_1^{k_1} \cdots p_m^{k_m}$, where the $p_i$'s are distinct primes. It turns +out that we can do something similar for groups. We first state the theorem; the +proof is difficult. + +\begin{theorem}[Classification of finite abelian groups] +\label{thm:classification-of-finite-abelian-groups} + Every finite abelian group is a unique product of cyclic groups of prime + power order. +\end{theorem} +Now let's see what this means. Let $G$ be a finite abelian group of order $n$. +Then, \cref{thm:classification-of-finite-abelian-groups} says that +\[ + G \cong \bZ_{p_1^{k_1}} \times \cdots \times \bZ_{p_m^{k_m}}. +\] +However, note that \emph{the $p_i$'s may not be distinct primes}. However, this +"factorization" is unique, meaning that if +\[ + G \cong \bZ_{q_1^{l_1}} \times \cdots \times \bZ_{q_n^{l_n}}, +\] +where $q_i$'s are primes, then $n = m$, and $\set{p_1, \dots, p_n} = \set{q_1, +\dots, q_m}$, and if $p_j = q_i$ then their powers are the same too. + + +This theorem is extremely powerful. It is extremely easy to determine \emph{all} +Abelian groups of a certain order. In contrast, classifying non abelian groups +is extremely difficult. We additionally obtain a partial converse to Lagrange's +theorem as a corollary. +\begin{corollary}[Subgroups of finite abelian groups] +\label{cor:subgroups-fin-abelian-groups} + Let $G$ be a finite abelian group and $m$ divide the order of $G$. Then $G$ + has a subgroup of order $m$. +\end{corollary} +\begin{proof} + See \cref{ex:subgroups-fin-abelian-groups}. +\end{proof} + +We will now prove the theorem. This proof comes from \autocite[Ch~11]{Gallian_2020}. + +\begin{lemma} + Let $G$ be a finite abelian group of order $p^n m$ where $p$ does not divide + $m$. Then, $G = H \times K$\footnote{This is an internal direct product. See + \cref{ex:internal-direct-products} for the definition of an internal direct + product.}, where $H = \set{x \in G: x^{p^n} = e}$ and $K = \set{x \in G: x^m + = e}$. Additionally, $\abs H = p^n$. +\end{lemma} +\begin{proof} + Obviously $H$ and $K$ are subgroups. Let us show that $H \cap K = \set{e}$ + and $HK = G$. +\end{proof} + + +\subsection{Exercises and Problems} + +\begin{exercise}[Subgroups of finite abelian groups] +\label{ex:subgroups-fin-abelian-groups} +Prove \cref{cor:subgroups-fin-abelian-groups}. A sketch is given in the next paragraph. + +Let $G$ be a finite abelian group of order $n$. We perform induction on $n$. +When $n=1$ or $m=1$ it is trivial. Suppose the theorem is true for all abelian +groups of order less than $n$. Let $p$ be a prime dividing $m$, so that $G$ has +a subgroup of order $p$, say $K$ (by Cauchy's Theorem). Then $G/K$ has order +$n/p$, and by the inductive hypothesis there is some subgroup of $G/K$ of the +form $H/K$ that has order $m/p$. +\end{exercise} + + +\end{document} \ No newline at end of file diff --git a/chapter-lagrange-theorem.tex b/chapter-lagrange-theorem.tex index 1b4820b..5b4531f 100644 --- a/chapter-lagrange-theorem.tex +++ b/chapter-lagrange-theorem.tex @@ -155,7 +155,8 @@ Prove \cref{cor:consequences-of-lagrange}. \end{exercise} -Let us now see an application of Lagrange's Theorem. +To really demonstrate the power of Lagrange's theorem, we shall see some +applications of it. The first application is in number theory. \begin{corollary}[Fermat's Little Theorem] \label{cor:fermat-little-theorem} @@ -172,4 +173,87 @@ Fill in the details of \cref{cor:fermat-little-theorem}. \end{exercise} +Lagrange's theorem also gives us a useful counting theorem which tells us what +the sizes of subgroups can be. +\begin{theorem}[$HK$ theorem] +\label{thm:hk-theorem} + Let $H, K$ be finite subgroups of some group $G$. Define $HK = \set{hk : h \in H, k \in K}$. Then, + \[ + \abs{HK} = \frac{\abs H \abs K}{\abs{H \cap K}}. + \] +\end{theorem} +Let's talk strategy. Of course, $HK$ has $\abs H \cdot \abs K$ products, but +they may not be distinct group elements. What this means is that we could have +$hk = h' k'$ where $h \neq h', k \neq k'$. The formula suggests that duplicates +occur in multiples of $\abs{H \cap K}$. We need some way to tie each product in +$HK$ to every single element of $\abs{H \cap K}$. The first observation comes +from noticing that if $t \in H \cap K$, then $hk = (ht)(t\inv k)$. + +\begin{proof} + Let $h \in H, k\in K$. If $t \in H \cap K$, then $hk = (ht)(t\inv k)$. This + tells us that every element of $HK$ is represented by at least $\abs{H \cap + K}$ products in $HK$. Suppose $hk = h' k'$, then, + \[ + t t\inv = h\inv h' k' k\inv. + \] + So if $t = h\inv h' = k {k'}\inv$, then it all works out. This shows that + every element in $HK$ is represented by precisely $\abs{H \cap K}$ products. +\end{proof} + + +Let us now see another application of Lagrange's theorem. This time, we classify +all groups of order $2p$ where $p$ is some odd prime. +\begin{theorem}[Classification of groups of order $2p$] +\label{thm:groups-order-2p} + Let $p$ be a prime such that $p > 2$. Let $G$ be a group of order $2p$. Then + $G$ is isomorphic to $\bZ_{2p}$ or $D_p$. +\end{theorem} +\begin{proof} + See \cref{ex:prove-thm-classifying-order-2p}. +\end{proof} + + +Counting can be useful. We now make use of Lagrange's theorem to prove a fact +about group actions. +\begin{theorem}[Orbit-Stabilizer Theorem] +\label{thm:orb-stab-thm} + Let $G$ be a finite group acting on a set $S$. Then, + \[ + \abs{G} = \abs{\operatorname{orb}_G (s)} \abs{\operatorname{stab}_G (s)}. + \] +\end{theorem} +\begin{proof} + The stabilizer of $s$ is a subgroup of $G$. It will suffice to provide a + bijection between left cosets of $\operatorname{stab}_G (s)$ and elements in + $\operatorname{orb}_G(s)$. The map $\varphi: g \operatorname{stab}_G (s) \mapsto g + \cdot s$ will do. We leave the details to the reader in + \cref{ex:orb-stab-thm}. +\end{proof} + + +\subsection{Exercises and Problems} + +\begin{exercise} +\label{ex:prove-thm-classifying-order-2p} + Prove \cref{thm:groups-order-2p} by following the steps below. + \begin{enumerate} + \item Assume that $G$ has no element of order $2p$. Show that $G$ must have an element of order $p$, call it $a$. + \item Find an element of order 2, call it $b$. + \item Show that $a$ and $b$ satisfy the relations of $D_p$: in particular, $a^j b = b a^{-j}$ for $j \in \set{1, \dots, p-1}$. + \item Show that every element of $G$ can be uniquely expressed in the form $a^jb^k$. + \item Conclude that $G$ is isomorphic to $D_p$. + \end{enumerate} +\end{exercise} + +\begin{exercise}[Orbit-Stabilizer Theorem] +\label{ex:orb-stab-thm} +Complete the proof of \cref{thm:orb-stab-thm}. In particular, show that the map +$\varphi$ as defined is well-defined and a bijection. Note that +the facts in \cref{ex:orbits-partition-set} will be needed. +\end{exercise} + +\begin{exercise} + Prove that the rotation group of a cube is $S_4$. +\end{exercise} + \end{document} \ No newline at end of file diff --git a/chapter-normal-subgroups-homomorphisms.tex b/chapter-normal-subgroups-homomorphisms.tex index f9dc95d..7dccdb9 100644 --- a/chapter-normal-subgroups-homomorphisms.tex +++ b/chapter-normal-subgroups-homomorphisms.tex @@ -28,11 +28,11 @@ \] \end{lemma} \begin{proof} - See \cref{ex:prove-normality0-criterion}. + See \cref{ex:prove-normality-criterion}. \end{proof} \begin{exercise} -\label{ex:prove-normality0-criterion} +\label{ex:prove-normality-criterion} Prove \cref{lem:normality-criterion} \end{exercise} @@ -323,7 +323,30 @@ \section{Isomorphism Theorems} and it has kernel $N$. -\section{Problems} +\section{Exercises and Problems} + +\begin{exercise} + Let $G$ be a group and let $x, y \in G$. Let $H$ be a subgroup of $G$. Show + that if $xH = Hy$, then $xHy\inv \subseteq H$. Find an example where they + are not equal. +\end{exercise} + +\begin{exercise}[Internal direct products] +\label{ex:internal-direct-products} + Let $G$ be a group. We say that $G$ is the \textbf{internal direct product + of $H$ and $K$} and write $G = H \times K$ if + \begin{enumerate} + \item $H, K$ are normal subgroups of $G$, + \item $HK = G$, + \item $H \cap K = \set{e}$. + \end{enumerate} + Note that this seemes similar to a vector space being a direct sum of + subspaces. + + Show that $G$ is isomorphic to $H \times K$. This justifies the abuse of the + group product notation. Extend this to the case where $G$ is an internal + direct product of a finite number of groups. +\end{exercise} \begin{exercise}[Center is always normal] \label{ex:center-is-always-normal}