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Solution_to_Algebra_Chapter_0.tex
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% Compile with PdfLaTeX
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\begin{document}
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\rightline{Hooy}
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%\name{Hooy}
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\vspace*{1.5em}
\problemlist{\textbf{\LARGE Algebra, Chapter 0}\\\vspace*{0.6em}By Paolo Aluffi}
\tableofcontents
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\appendix
\newpage
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\section*{Notation for Problems}
$\vartriangleright$: those problems that are directly referenced from the text.\\
\noindent $\neg$: those problems that are referenced from other
problems.\\
\noindent [\textsection II.8.1]: related to the text in II.8.1 (Chapter II Section 8 Subsection 1).\\
\noindent [II.8.10]: related to the Definition/Example/Proposition/Lemma/Corollary/Claim 8.10 in Chapter II (the 10th Definition/Example/Proposition/Lemma/Corollary/Claim in Chapter II Section 8).
\section*{Acknoledgement}
It is kind of Shane Creighton-Young to share his solutions to Paolo Aluffi's ``Algebra: Chapter 0"\cite{aluffi2009algebra} on the Github website \href{https://github.com/srcreigh/aluffi}{\texttt{https://github.com/srcreigh/aluffi}}. He takes the credit for some parts of the first two chapters of this manuscript.
\section*{Contact}
Github:
\href{https://github.com/hooyuser/Solution-to-Algebra-Chapter-0}{\texttt{https://github.com/hooyuser/Solution-to-Algebra-Chapter-0}}
\noindent E-mail: \href{mailto:hooyuser@outlook.com}{\texttt{hooyuser@outlook.com}}
\section*{License}
This work is licensed under a \href{http://creativecommons.org/licenses/by-nc-sa/4.0/}{Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License}.
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\include{chapter8.tex}
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\section{Appendix}
\hypertarget{Lemma II.1}{}
\textbf{Lemma II.1} (von Dyck)
Given a presentation $(A|\mathscr{R})=F(A)/R$, where $A$ is the set of generators, $\mathscr{R}\in F(A)$ is the set of relators and $R$ is the smallest normal subgroup of $F(A)$ containing $\mathscr{R}$. Define inclusion mapping $i:A\to F(A)$ and projection $\pi:F(A)\to F(A)/R$. If $f$ is a mapping from $A$ to a group $G$, and every relations in $\mathscr{R}$ holds in $G$ via $f$, that is, $\mathscr{R}\subset\ker\varphi$ where $\varphi$ is the unique homomorphism induced by the universal property of free group, then there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $f=\psi\circ\pi\circ i$. If $G$ is generated by $f(A)$, then $\psi$ is surjective.
\[\xymatrix{
F(A)/R\ar@{-->}[rd]^{\exists!\psi}\\
F(A)\ar@{-->}[r]^{\varphi}\ar[u]^{\pi} &G\\
A\ar[ru]_{f}\ar[u]^{i}&
}\]
\textbf{Proof of the lemma.} Since $R$ is the smallest normal subgroup of $F(A)$ containing $\mathscr{R}$ and the normal subgroup $\ker\varphi$ contains $\mathscr{R}$, we must have $R\subset\ker\varphi$. Then according to Theorem 7.12, there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $\varphi=\psi\circ\pi$, which means the whole diagram commutes. If there exists a homomorphism $\zeta:F(A)/R\to G$ such that $f=\zeta\circ\pi\circ i$, then we have $\varphi\circ i=\zeta\circ\pi\circ i$, which implies $\varphi(t)= \zeta(\pi(t))$ for all $t\in A$. Note that a homomorphism defined on $F(A)$ can be specified only by its valuation on the set of generators $A$, we can assert that $\varphi=\zeta\circ\pi$. Since there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $\varphi=\psi\circ\pi$, we have $\zeta=\psi$. Thus we show that there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $f=\psi\circ\pi\circ i$.
Moreover, if $G$ is generated by $f(A)$, then $\mathrm{im}\psi=G$, since $f(A)=\psi(\pi( i(A)))\subset\mathrm{im}\psi$ implies $G\subset\mathrm{im}\psi$.\hfill$\lrcorner$
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