\documentclass[11pt]{article} % Define the document type. The 11pt refers to font size.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Include packages required to compile document:

\usepackage{amssymb, amsthm, amsmath} % These packages are used for the basic math symbols and environments.
\usepackage{enumerate} % This package is used to provide various types of numbering for lists.
\usepackage[margin=1.25in]{geometry} % This is to set wide margins, so there is room to leave comments
\usepackage[parfill]{parskip} % This provides formatting for paragraphs so that there are no indents and paragraphs are separated by a line skip.
\usepackage[pdftex,colorlinks=true,urlcolor=blue]{hyperref}
% You may need to add additional packages depending on the commands you want to use.

%%%%%%%%%%%%%%%%%%%%
% User defined commands:

\newcommand{\br}{\mathbb R} % this is the symbol for the real numbers.  mathbb is the "blackboard bold" font
\newcommand{\bz}{\mathbb Z} % integers
\newcommand{\bq}{\mathbb Q} % rationals
\newcommand{\bn}{\mathbb N} % natural numbers

\newcommand{\co}{\colon\thinspace} % this is a good way to write functions, i.e instead of $f: X \to Y$, you would write $f\co X \to Y$.  It looks nicer.
\newcommand{\ssm}{\smallsetminus} % A nice way of writing set subtraction 

% You may add shortcuts for any commands you find yourself using frequently.  

\DeclareMathOperator{\ord}{ord} % This command lets you declare math operators.  Some built in operators are \sin for sine and \cos for cosine.  This makes the operators appear nicely in the math environment.  If I wanted to make the sine operator myself, it would look like \DeclareMathOperator{\sin}{sine}.  

% Ender user defined commands
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% These establish different environments for stating Theorems, Lemmas, Remarks, etc.

\newtheorem{Thm}{Theorem}
\newtheorem{Prop}[Thm]{Proposition}
\newtheorem{Lem}[Thm]{Lemma}
\newtheorem{Cor}[Thm]{Corollary}

\theoremstyle{definition}
\newtheorem{Def}[Thm]{Definition}

\theoremstyle{remark}
\newtheorem{Rem}[Thm]{Remark}
\newtheorem{Ex}[Thm]{Example}

\theoremstyle{definition}
\newtheorem{Exercise}{Exercise}
\newtheorem{ExGraded}[Exercise]{*Exercise}
\newtheorem{ExChallenge}[Exercise]{**Exercise}

\newenvironment{Solution}{\noindent{\it Solution.}}{\qed}

%End environments
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Now we're ready to start
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

% This is for the title
{\Large {\bf Homework 10}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
\line(1,0){470}

\medskip





\begin{Exercise}[Section 9.4 \#22]
Let \( G \) be a group of order 20. 
If \( G \) has subgroups \( H \) and \( K \) of orders 4 and 5, respectively, such that \( hk = kh \) for all \( h \in H \) and \( k \in K \), prove that \( G \) is the internal direct product of \( H \) and \( K \). 
\end{Exercise}

\begin{Solution}
Since we are given that every element of \( H \) commutes with every element of \( K \), we only need to check that \( H \cap K = \{e \} \) and \( G = HK \). 
Let us first show that the intersection is trivial.
If \( a \in H \cap K \), then by Lagrange's Theorem, the order of \( a \) must divide the order of \( H \) and also the order of \( K \).
As the orders of \( H \) and \( K \) are relatively prime, we must have that \( |a| = 1 \), and hence \( a = e \); in particular, \( H \cap K = \{e\} \).

Now, the set \( HK = \{ hk : h \in H, k \in K \} \) has at most \( |H|\cdot |K| \) elements; we claim it has exactly that many elements.
Suppose \( h, h' \in H \) and \( k, k' \in K \) such that \( hk = h'k' \).
Rearranging, this tells us that \( (h')^{-1}h = k' k^{-1} \).
The product on the left-hand side is in \( H \) and the product on the right-hand side is in \( K \). 
But \( H \cap K =\{e\} \), so we must have that \( (h')^{-1}h = k'k^{-1} = e \), implying that \( h' = h \) and \( k' = k \). 
This establishes that \( |HK| = |H|\cdot |K| \).
Therefore, as \( |H| \cdot |K| = |G| \), we have that \( |G| = |HK| \), implying \( G = HK \).
We have now shown that \( G \) is the internal direct product of \( H \) and \( K \). 
\end{Solution}

\setcounter{Exercise}4

\medskip
\begin{ExChallenge}
Let \( N \) be a group, and let \( H \) be a subgroup of \( \rm{Aut}(N) \), the automorphism group of \( N \).
The \emph{(external) semidirect product} of \( N \) and \( H \) is the group \( N \rtimes H \) whose underlying set is \( N \times H \) and whose group operation is defined by \( (a, \varphi)(b, \psi) = (a\varphi(b), \varphi \circ \psi) \). 
\begin{enumerate}[(a)]
\item Prove that \( N\rtimes H \) is a group (yes, I said it was a group in the definition, but that needs a proof).
\item Let \( G \) be a group, and let \( N \) and \( H \) be subgroups of \( G \) such that \
\begin{enumerate}[(i)]
\item \( N \cap H = \{e\} \),
\item  \( G = NH = \{nh : n \in N, h \in H \} \), and
\item  \( h n h^{-1} \in N \) for all \( n \in N \) and all \( h \in H \).
\end{enumerate}
(Note that condition (ii) and (iii) together imply that \( N \) is a normal subgroup of \( G \), see the Week 12 notes.)
Condition (iii) says that each element of \( H \) induces an automorphism of \( N \) via conjugation, that is, for \( h \in H \) we can define \( \varphi_h \in \rm{Aut}(N) \) by \( \varphi_h(n) = hnh^{-1} \) for all \( n \in N \). 
Identifying \( h \) with \( \varphi_h \), we can view \( H \) as a subgroup of \( \rm{Aut}(N) \). 

Under these hypotheses and under this identification of \( H \) with a subgropup of \( \rm{Aut}(N) \), prove that \( G \cong N \rtimes H \). 
Here, we say \( G \) is the \emph{(internal) semidirect product} of \( N \) and \( H \).

\end{enumerate}
\end{ExChallenge}

\begin{Solution}
\begin{enumerate}[(a)]

\item Associativity follows from the associativity of the group operations of \( N \) and \( H \); I'll leave the details to you.
There is an identity element, namely \( (e_N, e_H) \). 
It is left to check inverses. 
If \( (a, \varphi) \in N \rtimes H \), then observe that \( (a, \varphi)(\varphi^{-1}(a^{-1}), \varphi^{-1}) = (e_N, e_H) \), so \( (a,\varphi)^{-1} = (\varphi^{-1}(a^{-1}), \varphi^{-1}) \). 

\item
Define \( \Phi \co N \rtimes H \to G \) by \( \Phi(n,h) = nh \). 
Condition (i) tells us that \( \Phi \) is injective, and condition (ii) tells us that \( \Phi \) is surjective. 
To finish, let \( (n, h), (n',h') \in N\rtimes H \). 
Then
\begin{align*}
 \Phi((n,h)(n',h')) 	&= \Phi(nhn'h^{-1}, hh') \\
 				&= nhn'h^{-1}hh' \\
				&= nhn'h' \\
				&= \Phi(n,h)\Phi(n',h').
\end{align*}
All together, we have that \( \Phi \) is an isomorphism. 
\end{enumerate}
\end{Solution}

\end{document}