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\begin{document}

\title{Introduction to algebra --- handout 8}
\maketitle

Let us note that in commutative groups every subgourp is normal. 

\begin{thm}
The lattice of normal subgroups is modular.
\end{thm}

\begin{proof}
The normal subgropus form a sublattice of the lattice of subgroups, 
because (1) the intersection of normal subgrpoups is normal and (2)
a subgroup generated by two normal subgrpous is normal. 

To show the modularity, let $L\subseteq K$ be two normal subgroups.
We have to show that:
\begin{equation*}
L\lor (H\land K)\supseteq (L\lor H)\land K
\end{equation*}

Because $L$ and $H$ are normal subgroups, $L\lor H=LH$. Therefore, we
have to show that:
\begin{equation*}
L(H\cap K)\supseteq (LH)\cap K
\end{equation*}
Let $g\in (LH)\cap K$. This means that $g\in K$ and ther is $l\in L$
and $h\in H$ such that $g=lh$. $h=l^{-1}g\in LK\subseteq K^2=K$ since
$g\in K$. Therefore, $h\in (H\cap K)$. Therefore, $g=lh\in L(H\cap
K)$.
\end{proof}

\begin{prop}
If $N\triangleleft G$, then the set of left cosets $G/N$ form a group, called the {\bf quotient group}. 
\end{prop}

\begin{ex}
$H=\langle m\Z, +\rangle\triangleleft \langle \Z, +\rangle=G$ and $Z_m=G/H$.
\end{ex}


\begin{df}
A map $\phi$ from group $\langle G, \cdot\rangle$ to group $\langle H,
*\rangle$ is called {\bf group homomorphism} if $\phi(a\cdot
b)=\phi(a)*\phi(b)$. A bijective homomorphism is called {\bf isomorphism}. 
\end{df}

\begin{df}
Let $\phi$ be a group homomorphism from $G$ to $H$. The {\bf kernel} of
$\phi$ is the set of those elements of $G$ which are mapped into the
identity of $H$:
\begin{equation*}
ker(\phi):=\{ g\in G : \phi(g)=e_H\}. 
\end{equation*}
The {\bf image} of $\phi$ is the set of those elements of $H$ which are the image of some elements of $G$:
\begin{equation*}
im(\phi)=\{ h\in H : h=\phi(g) \text{ for some } g\in G\}.  
\end{equation*}
\end{df}

\begin{e}[first isomorphism theorem]
Let $\phi$ be a group homomorphism from $G$ to $H$. 
Show that $ker(\phi)$ is a normal subgroup of $G$, $im(\phi)$ is subgroup of $H$ and $im(\phi)\cong G/ker{\phi}$. 
\end{e}

\begin{df}
Let $X$ be a set. Permutations of $X$ form a group with respect to
composition. This group is called the {\bf symmetric group on $X$} and
denoted by $Sym(X)$.
\end{df}

\begin{thm}[Cayley] 
Every group $G$ is isomorphic to a subgroup of $Sym(G)$. 
\end{thm}

\begin{proof}
Let $g\in G$. Consider map $f_g:G\rightarrow G$, $f_g(x)=gx$. $f_g$ is
a permutation of set $G$ since it is surjective [$f_g(g^{-1}y)=y$] and
invertible [$f^{-1}_g=f_{g^{-1}}$].  The map $\phi:G\ni g\mapsto
f_g\in Sym(G) $ is an injective group homomorphism because
$f_{ab}=f_{a}\circ f_{b}$ and, if $f_a=f_b$,
$a=f_a(e)=f_b(e)=b$. Consequently, $\phi:G\rightarrow im(\phi)\le
Sym(G)$ is an isomorphism.
\end{proof}
\end{document}