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\begin{document}

\title{Introduction to algebra --- handout 9}
\maketitle

\begin{prop}
Let $G$ and $H$ be two groups. Then $G\times H$ is a group with 
\begin{equation*}
(g_1,h_1)\cdot(g_2,h_2):=(g_1\cdot g_2,h_1\cdot h_2).
\end{equation*}
This group is called the  {\bf direct product} of $G$ and $H$. 
\end{prop}

\begin{proof}
It is straightforward to show the associativity. The identity of $G\times H$ is $(e_g,e_h)$, and the inverse of $(g,h)$ is $(g^{-1},h^{-1})$.  
\end{proof}

\begin{ex} 
 $\R\times \R$, $\Z_2\times \Z_2$ or $\Q\times \Z_2$. 
\end{ex}

\begin{prop}\label{int}
$G'=\{(g,e_h): g\in G\}$ and $H'=\{(e_g,h): h\in H\}$ are two subgroups of $G\times H$ isomorphic to $G$ and $H$, and they  have the following properties 
\begin{enumerate}
%\item[(0)] $G\cong G'$, $H\cong H'$;
\item $G'\cap H'=\{e\}$;
\item $G\times H=G' H'$, i.e., for all $a\in G\times H$, there are $g\in G'$ and $h\in H'$ such that $a=gh$;  
\item $G'\triangleleft G\times H$ and $H'\triangleleft G\times H$ (or equivalently $gh=hg$ for all $g\in G'$ and $h\in H'$);
\end{enumerate}
\end{prop}

\begin{df}
If two subgroups $N_1$ and $N_2$ of a group $G$ have the properties of Proposition~\ref{int}, then $G$ is called the {\bf internal direct product} of $N_1$ and $N_2$.   
\end{df}


\begin{lem}
$\Z_{mn}$ is isomorphic to $\Z_m\times\Z_n$ iff $m$ and $n$ are
  relative primes, i.e., their greatest common divisor is $1$.
\end{lem}

\begin{ex} ${}$
\begin{itemize}
\item $\Z_6\cong \Z_2\times \Z_3$. 
\item $\Z_2\times\Z_2\not\cong \Z_4$.
\end{itemize}
\end{ex}

\begin{thm}
Every finitely generated abelian group can be decomposed as
\begin{equation*}
\Z^n\times \Z_{p^{n_1}_1}\times\ldots \times\Z_{p^{n_k}_k},
\end{equation*}
where $p_1$, \ldots, $p_k$ are prime numbers. This decomposition is unique up to ordering.
\end{thm}

\begin{ex} ${}$
\begin{itemize}
\item $\Z_{42}\cong \Z_2\times\Z_3\times\Z_7$, 
\item $\Z_{12}\cong \Z_3\times\Z_4\not\cong\Z_3\times\Z_2\times\Z_2$
\end{itemize}
\end{ex}


\begin{ex}
$\langle\Q,+\rangle$ is not finitely generated. 
\end{ex}

\begin{df}
A group $G$ is called {\bf Boolean group} if $a^2=e$ for all $a\in G$.  
\end{df}


\begin{e} ${}$ Show that
\begin{itemize}
\item every Boolean group is commutative, and 
\item every finite Boolean group has $2^n$ elements for some natural number $n$.  
\end{itemize}
\end{e}

\begin{prop}\label{BA}
Let $\langle B, \land, \lor, \bar~ , 0, 1\rangle$ be a Boolean algebra. Then $B$ is a Boolean group with the symmetric difference $x\triangle y:=(\bar x \land y)\lor (x\land \bar y)$.  
\end{prop}

\begin{proof}
Associativity: calculation. Identity is $0$ and $x\triangle x=0$. 
\end{proof}


\begin{df}
Theory $T$ said to be {\bf interpretable} in $S$ if every basic concept of $T$ can be defined explicitly in $S$ such that the translation of the axioms of $T$ are theorems in $S$.  
\end{df}

\begin{ex}
By Proposition~\ref{BA}, the theory of Boolean groups is interpretable in the theory of Boolean algebras.  
\end{ex}

\begin{prop}
The theory of Boolean algebras is not interpretable in the theory of Boolean Groups. 
\end{prop}


\end{document}