Skip to content

Commit

Permalink
add hk theorem, groups of order 2p theorem and orb stab theorem
Browse files Browse the repository at this point in the history
  • Loading branch information
bushshrub committed Sep 2, 2024
1 parent 43a15e6 commit 82a2d68
Show file tree
Hide file tree
Showing 5 changed files with 204 additions and 4 deletions.
3 changes: 3 additions & 0 deletions .vscode/settings.json
Original file line number Diff line number Diff line change
@@ -0,0 +1,3 @@
{
"makefile.configureOnOpen": false
}
9 changes: 9 additions & 0 deletions chapter-4-permutation-groups.tex
Original file line number Diff line number Diff line change
Expand Up @@ -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}
81 changes: 81 additions & 0 deletions chapter-classification-of-finite-abelian-groups.tex
Original file line number Diff line number Diff line change
@@ -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}
86 changes: 85 additions & 1 deletion chapter-lagrange-theorem.tex
Original file line number Diff line number Diff line change
Expand Up @@ -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}
Expand All @@ -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}
29 changes: 26 additions & 3 deletions chapter-normal-subgroups-homomorphisms.tex
Original file line number Diff line number Diff line change
Expand Up @@ -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}

Expand Down Expand Up @@ -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}
Expand Down

0 comments on commit 82a2d68

Please sign in to comment.