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add hk theorem, groups of order 2p theorem and orb stab theorem
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{ | ||
"makefile.configureOnOpen": false | ||
} |
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\documentclass[./main.tex]{subfiles} | ||
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\begin{document} | ||
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\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. | ||
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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. | ||
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\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. | ||
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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} | ||
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We will now prove the theorem. This proof comes from \autocite[Ch~11]{Gallian_2020}. | ||
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\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} | ||
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\subsection{Exercises and Problems} | ||
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\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. | ||
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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} | ||
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\end{document} |
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