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\begin{document}
 \parindent=0cm
  \section*{III.1 Amalgamated product }

\begin{defn}$G, G_1 , G_2 :$ groups with $f_i : G \rightarrow G_{i} $, homomorphism for $ i= 1 ,2$.\\
$F$ :=  the free group generated by $G_1 \amalg G_2$. Denote by
 $x\cdot y$ the product in $F$.\\
$R$ := the normal subgroup of $F$ generated by the words
$(xy)\cdot y^{-1} \cdot x^{-1}$,\\ $\hspace*{2em}$where both
$x,y\in G_i\hspace{0.5em},\hspace{0.5em} i =1,2$ and $f_1 (z)
\cdot f_2 (z)^{-1}$ for $z\in G$.

The amalgamated product of $G_1$ and $G_2$ over G, $G_1
\underset{G}{*} G_2 := F/R$\end{defn}

\textbf{Remark.}  $ G_1 * G_2 := G_1 \underset{\{1\}}{*} G_2 $\\

\textbf{Note.} $\hspace{3em}G_1\\
\hspace*{3em}f_1 \nearrow \hspace{3em} \searrow g_1 \\
\hspace*{2em}G\hspace{3em} \curvearrowright \hspace{2em}G_1
\underset{G}{*} G_2\\
\hspace*{3em}f_2 \searrow \hspace{3em} \nearrow g_2 \\
\hspace*{6em}G_2$\\

where the canonical map $g_{i} : G_{i}\overset{i}{\hookrightarrow}
F\overset{p}{\rightarrow} F/R = G_1 \underset{G}{*} G_2$ is a homomorphism.\\

Then $g_1  f_1 (z) = g_2  f_2 (z)$ by the definition of  $F/R$.\\

\textbf{2. universal property of amalgamation}\\

(a) Suppose $h_{i} : G_{i} \rightarrow H$ is a homomorphism with
$h_1 f_1 = h_2
f_2.$\\
Then $\exists {!}$ homomorphism $h : G_1 \underset{G}{*} G_2
\rightarrow H$ such that $hg_{i} = h_{i}\hspace{0.5em},\hspace{0.5em}i = 1,2$ .\\

(b) If $H$ is generated by $h_1 (G_1 ) \cup h_2 (G_2 )$, then $h$
is
onto.\\
\framebox{\hspace*{1em}\parbox[b]{14cm}{ $\hspace*{6em}G_1\\
\hspace*{3em}f_1\nearrow \hspace{3em}$\textcolor{blue}{$ \searrow g_1$}$\hspace{1em}\searrow h_1 \\
\hspace*{2em}G\hspace{3em} \curvearrowright
\hspace{1em}$\textcolor{blue}{$G_1
\underset{G}{*} G_2\overset{h}{\rightarrow}$}$H\\
\hspace*{3em}f_2 \searrow \hspace{3em} $\textcolor{blue}{$\nearrow g_2$}$\hspace{1em}\nearrow h_2 \\
\hspace*{6em}G_2$}}\\

\begin{proof}\\
(a) Define $h' : F \rightarrow H$ as the unique homomorphism determined by the condition $h' |_{G_{i}} = h_{i}.$\\
\newpage
$R \subset$ ker$(h')$:\\

$R$��  generator $(xy)\cdot y^{-1} \cdot x^{-1}$,$f_1 (z)\cdot f_2
(z)^{-1}$�� ���ؼ��� check�ϸ� �ȴ�.\\
$h_1 (xy) h_1 (y^{-1} ) h_1(x^{-1} ) = 1$ for $x,y \in G_1$,\\
$h_2 (xy) h_2 (y^{-1} ) h_2(x^{-1} ) = 1 $ for $x,y \in G_2$ and\\
$h_1 (f_1 (z))h_2 (f_2 (z) ^{-1} ) = h_1 f_1 (z) (h_2 f_2
(z))^{-1} = 1$ ($\because h_1 f_1 = h_2 f_2 $)\\

$\therefore \exists h : F/R \rightarrow H.$ \\

$G_{i}$ �󿡼� $h' = h_{i}\Rightarrow h_{i} = h' i = hpi =h
g_{i}\Rightarrow hg_{i} = h_{i}$\\

$\hspace*{1em} G_{i} \hspace{0.5em}\overset{i}{\hookrightarrow}
 F \overset{h' }{\rightarrow}\hspace{0.5em}
H\\
\hspace*{1em}g_i \searrow \curvearrowright \downarrow
p \curvearrowright \nearrow h$ : diagram commutes from definitions of $g_{i}$ and {h}.\\
$\hspace*{3.1em} G_1 \underset{G}{*} G_2\\
\hspace*{3.4em} F/R$\\

Uniqueness is obvious since $h$ is already determined on $g_{i}
(G_{i} )$ \\and $g_1 (G_1 )\cup g_2 (G_2 )$ generates $G_1
\underset{G}{*} G_2$.\\

(b) Suppose $\forall a \in H$ is $ a = a_1 a_2 \cdots a_k$.\\
$a_j \in h_1 (G_1 ) \cup h_2 (G_2 )\Rightarrow a_{j} = h_{i}
(x_{j} )$ for $ i = i(a_j ) = $ 1 or 2\\
$\Rightarrow a = \Pi a_{j} = \Pi h_{i} (x_{j} ) = \Pi hg_{i}
(x_{j} ) = h ( \Pi g_{i} (x_{j} ))$\end{proof}
\\

\textbf{���� 9.}\\
(1) $G_1 \underset{G}{*} G_2 \cong G_1 * G_2 / \Gamma$,\\
$\hspace*{1em}$ where $\Gamma$ is the normal subgroup generated by
$f_1
(z)\cdot f_2(z) ^{-1}$.\\
(2) $f_1$ : onto $\Rightarrow \hspace{1em}g_2 :$ onto\\
(3) $f_1$ : $\cong \hspace{1em} \Rightarrow \hspace{1em} g_2$ :
$\cong$\\
\newpage
\textbf{3. Group presentation.}
\begin{thm}
$G_1 = <x_1 , \cdots ,x_{n} | r_1 , \cdots ,r_{k} >\\
\hspace*{3.5em}G_2 = <y_1 , \cdots ,y_{m} | s_1 ,\cdots ,s_{l} >$\\
$\hspace*{17em} f_1 \nearrow G_1$\\
$\hspace*{3.5em}G = <z_1 ,\cdots ,z_{p} | $ any $>$ with G \\
$\hspace*{17em}f_2\searrow G_2$\\
 $\Rightarrow G_1 \underset{G}{*}
G_2 \cong <x_1 ,\cdots ,x_{n} ,y_1 , \cdots ,y_{m} | r_1 ,\cdots
,r_{k} , s_1 ,\cdots ,s_{l}$ and $f_1 (z_{i} )f_2 (z_{i} )^{-1}$\\
$\hspace*{5em}$ for $i = 1 , \cdots , p>$ \end{thm}
\begin{proof}
Let $\overline{G} = <x_1 ,\cdots ,x_{n} ,y_1 , \cdots ,y_{m} | r_1
,\cdots
,r_{k} , s_1 ,\cdots ,s_{l}$ and $f_1 (z_{i} )f_2 (z_{i} )^{-1}$\\
$\hspace*{7em}$ for $i = 1 , \cdots , p>$.\\
Define $\phi : \overline{G} \rightarrow G_1 \underset{G}{*} G_2 $
by\\
$\hspace*{5em} x_{i} \mapsto g_1 ("x_{i} " )\hspace{1em} ( "x_{i} " \in G_1 )\\
\hspace*{5em} y_{i} \mapsto g_2 ("y_{i} " )\hspace{1em} ( "y_{i} " \in G_2 )$\\

$r_{i} = x_{i_1 } \cdots x_{i_{k} }$ in $\overline{G} \Rightarrow
\phi (r_{i} ) = \phi (x_{i_1 } \cdots x_{i_{k} } )= g_1 ("x_{i_1}
" ) \cdots g_1 ("x_{i_{k}}")\\ \hspace*{12em} = g_1 ("x_{i_1}
\cdots x_{i_{k}}")
= g_1 ("r_i " ) \\ \hspace*{12em}= g_1 (1) = 1$\\
Similarly $\phi (s_{j} ) =1$\\

$\phi ( f_1(z_{i}) f_2(z_{i} )^{-1} ) = g_1 (f_1 (z_{i} )) g_2
(f_2
(z_{i} ))^{-1} = 1$\\

$\Rightarrow \phi $ is a homomorphism.\\

For the converse, use universal property:\\

�Ʒ� diagram���� $h_1 (x_{i} ) = x_{i} , h_2 (y_{j} ) = y_{j}$��
�θ� $\overline{G}$�� ���ǿ� ���� $h_1 f_1 (z_{i} ) = h_2 f_2
(z_{i})$�� �����ϰ� ���� universal property�� ���� \\
$\hspace*{15em}G_1\\
\hspace*{12em}f_1\nearrow \hspace{3em}$\textcolor{blue}{$ \searrow g_1$}$\hspace{1em}\searrow h_1 $\\
$\exists {!} h: G_1 \underset{G}{*} G_2 \rightarrow \overline{G}$
s.t. $\hspace{0.5em}G\hspace{3em} \curvearrowright
\hspace{1em}$\textcolor{blue}{$G_1
\underset{G}{*} G_2\overset{h}{\rightarrow}$}$\overline{G} \hspace{2em}$ commute.\\
$\hspace*{12em}f_2 \searrow \hspace{3em} $\textcolor{blue}{$\nearrow g_2$}$\hspace{1em}\nearrow h_2 \\
\hspace*{15em}G_2$\\

Now $h\phi (x_{i} ) = h g_1(x_{i} ) = h_1(x_ {i}
)=x_{i}$ and \\
$h \phi (y_{j} ) = h g_2(y_{j} ) = h_2(y_
{j} )=y_{j} \\\Rightarrow \hspace{1em} h \phi =$id.\\

$\phi h(g_1 (x_{i} )) = \phi h_1 (x_{i} ) = \phi(x_{i} ) = g_1
(x_{i} )$ and\\ $\phi h(g_2 (y_{j} )) = \phi h_2 (y_{j} ) =
\phi(y_{j} ) = g_2 (y_{j} )\\\Rightarrow \hspace{1em} \phi h
=$id.\\

$\therefore \phi$ is an isomorphism with inverse $h$.
\end{proof}
\\
\\
\textbf{4. Special Case}\\

$G_1 = \{1\}$ (or $G_2 = \{1\}) \Rightarrow G_1 \underset{G}{*}
G_2 = G_2 / < f_2 (G) >$\footnote{ $<f_2(G)>$ is the normal
subgroup
generated by $f_2 (G)$}{}\\
\\
\end{document}
