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101 lines
3.3 KiB
101 lines
3.3 KiB
\documentclass{article}
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\usepackage[utf8]{inputenc}
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\usepackage{amsfonts}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{enumerate}
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\usepackage{hyperref}
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\hypersetup{
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colorlinks,
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citecolor=black,
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filecolor=black,
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linkcolor=black,
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urlcolor=black
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}
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% custom solution environment to set custom numbers
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\theoremstyle{definition}
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\newtheorem{innersolution}{Solution}
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\newenvironment{solution}[1]
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{\renewcommand\theinnersolution{#1}\innersolution}
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{\endinnersolution}
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\title{Seminar exercises}
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\author{ }
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\date{February 2022}
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\begin{document}
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\maketitle
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\begin{solution}{1.9}\
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\begin{enumerate}[1.]
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\item Let $f(a) = u$, then $g(f(a)) = g(u)$, so $g \circ f$ is a function.
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\item We can see that composition of functions is associative as follows:\\
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we know that $[ f \circ g](x) = f(g(x)), \forall x \in A$,\\
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so,
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$$(h \circ [g \circ f])(x) = h([g \circ f](x)) = h(g(f(x)))$$
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\\
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and
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$$([h \circ g] \circ f)(x) = [h \circ g](f(x)) = h(g(f(x)))$$
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Then, we can see that $$h \circ (g \circ f) = h(g(f(x))) = (h \circ g) \circ f$$
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\end{enumerate}
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\end{solution}
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\begin{solution}{1.28}\
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\emph{(WIP)}\\
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It is isomorphic to the cosets of the \emph{nth} roots of unity, which are $\mathbb{G}_n = \{w_k\}^{n-1}_{k=0}$, where $w_k=e^{\frac{2 \pi i}{n}}$.
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\end{solution}
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\begin{solution}{2.2}\
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To prove that the inverse $x^{-1}$ is unique, assume $x^{-1}$ and $\tilde{x}^{-1}$ are two inverses of $x$.\\
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By the definition of the inverse, we know that $x \cdot x^{-1} = e$. And by the definition of the unit element, we know that $x \cdot e = x$.\\
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Then, $$x^{-1} \cdot (x \cdot \tilde{x}^{-1}) = x^{-1} \cdot e = x^{-1}$$
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and $$(x^{-1} \cdot x) \cdot \tilde{x}^{-1} = e \cdot \tilde{x}^{-1} = \tilde{x}^{-1}$$
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By associativity property of groups, we know that
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$$x^{-1} \cdot (x \cdot \tilde{x}^{-1}) = (x^{-1} \cdot x) \cdot \tilde{x}^{-1}$$
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so, $$x^{-1} \cdot e = e \cdot \tilde{x}^{-1}$$
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which is $$x^{-1} = \tilde{x}^{-1}$$
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So, for any $x \in G$, the inverse $x^{-1}$ is unique.
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\end{solution}
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\begin{solution}{2.5}\
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Let $\alpha = (\begin{smallmatrix}1 & 2 & 3\\ 1 & 3 & 2\end{smallmatrix})$, $\beta = (\begin{smallmatrix}1 & 2 & 3\\ 3 & 1 & 2\end{smallmatrix})$, then,
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$$
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\alpha \cdot \beta =
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(\begin{smallmatrix}1 & 2 & 3\\ 1 & 3 & 2\end{smallmatrix})
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\cdot (\begin{smallmatrix}1 & 2 & 3\\ 3 & 1 & 2\end{smallmatrix})
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= (\begin{smallmatrix}1 & 2 & 3\\ 3 & 2 & 1\end{smallmatrix})
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$$
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and
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$$
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\beta \cdot \alpha =
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(\begin{smallmatrix}1 & 2 & 3\\ 3 & 1 & 2\end{smallmatrix}) \cdot
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(\begin{smallmatrix}1 & 2 & 3\\ 1 & 3 & 2\end{smallmatrix})
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= (\begin{smallmatrix}1 & 2 & 3\\ 2 & 1 & 3\end{smallmatrix})
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$$
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So, we can see that
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$$
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(\begin{smallmatrix}1 & 2 & 3\\ 3 & 2 & 1\end{smallmatrix})
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\neq
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(\begin{smallmatrix}1 & 2 & 3\\ 2 & 1 & 3\end{smallmatrix})
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$$
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so, $\alpha \cdot \beta \neq \beta \cdot \alpha$.
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\end{solution}
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\begin{solution}{2.26}\
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We want to prove that $f: G \rightarrow H$ is a \emph{monomorphism} iff $\ker f=\{e\}$.\\
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We know that $f$ is a \emph{monomorphism} (\emph{injective}) iff $\forall a, b \in G$, $f(a) = f(b) \Rightarrow a = b$.\\
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Let $a, b \in G$ such that $f(a)=f(b)$. Then
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$$f(a) f(b)^{-1} = f(b) (f(b))^{-1} = e$$
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$$f(a) f(b^{-1}) = e$$
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$$f(ab^{-1}) = e$$
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as $\ker f = \{e\}$, then we see that $ab^{-1}=e$, so $a=b$. Thus $f$ is a \emph{monomorphism}.
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\end{solution}
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\end{document}
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