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\def\R{{\mbox{$\mathbb R$}}}
\def\NN{{\mbox{$\mathbb N$}}}
\def\Z{{\mbox{$\mathbb Z$}}}
\def\Q{{\mbox{$\mathbb Q$}}}
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\def\C{{\mbox{$\mathbb C$}}}
\def\E{{\mbox{$\mathcal{E}$}}}
\def\M{{\mbox{$\mathcal{M}$}}}
\def\N{{\mbox{$\mathcal{N}$}}}

\def\A{{\mbox{$\mathcal{A}$}}}
\def\O{{\mbox{$\mathcal{O}$}}}

\def\S{{\mbox{$\mathcal{S}$}}}
\def\L{{\mbox{$\mathcal{L}$}}}
\def\P{{\mbox{$\mathcal{P}$}}}
\def\B{{\mbox{$\mathcal{B}$}}}
\def\X{{\mbox{$\mathcal{X}$}}}
\def\V{{\mbox{$\mathcal{V}$}}}
\def\Y{{\mbox{$\mathcal{Y}$}}}
\def\W{{\mbox{$\mathcal{W}$}}}
\def\dist{{\mbox{\rm dist}}}
\def\span{{\mbox{\rm span}}}
\def\ecc{{\mbox{\rm ecc}}}
\def\max{{\mbox{\rm max}}}
\def\spec{{\mbox{\rm spec}}}
\def\rank{{\mbox{\rm rank}}}
\def\mod{{\mbox{\rm mod}}}

\def\lcm{{\mbox{\rm lcm}}}

\def\id{{\mbox{\rm id}}}
\def\div{{\mbox{\rm div}}}
\def\trag{{\mbox{\rm trag}}}
\def\tr{{\mbox{\rm tr}}}
\def\dim{{\mbox{\rm dim}}}
\def\diag{{\mbox{\rm diag}}}
\def\Mat{{\mbox{\rm Mat}}}
\def\ker{{\mbox{\rm ker}}}
\def\im{{\mbox{\rm im}}}
\def\gcd{{\mbox{\rm gcd}}}

\def\J{{\mbox{$\pmb{J}$}}}
\def\I{{\mbox{$\pmb{I}$}}}
\def\j{{\mbox{$\pmb{j}$}}}
\def\0{{\mbox{\boldmath{$0$}}}}
\def\algmult{{\mbox{\rm alg\,mult}}}
\def\geomult{{\mbox{\rm geo\,mult}}}
\def\char{{\mbox{\rm char}}}
\def\du{{\mbox{\rm \,d$u$}}}
\def\dx{{\mbox{\rm \,d$x$}}}
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\def\dy{{\mbox{\rm \,d$y$}}}
\def\dz{{\mbox{\rm \,d$z$}}}
\def\tg{{\mbox{\rm tg\,}}}
\def\arctg{{\mbox{\rm arc\,tg\,}}}
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\begin{document}




\raggedright


\renewcommand{\thesection}{}
\section{\centerline{Algebra III - Abstraktna algebra, 13.06.2017.}}










\bigskip
\begin{zad}
Naj bo $S_{[0,1]}$ množica vseh bijekcij iz intervala $[0,1]$ na interval $[0,1]$.
\begin{itemize}
\item[{\rm (a)}] Pokaži, da je $S_{[0,1]}$ grupa glede na operacijo komponiranja preslikav.
\item[{\rm (b)}] Naj bo $T=\{\alpha\in S_{[0,1]}\,|\,\alpha(0)=0\}$. Pokaži, da je $T$ podgrupa grupe $S_{[0,1]}$ (razloži tudi, zakaj je $T$ neprazna množica).
\item[{\rm (c)}] Pokaži, da je $\{\sigma\in S_{[0,1]}\,|\,\sigma(0)=1\}$ levi odsek podgrupe $T$ v grupi $S_{[0,1]}$.
\item[{\rm (d)}] Določi $[S_{[0,1]}:T]$.
\end{itemize}
\end{zad}


\medskip
Re. 

\smallskip
(a) $\circ$ je asocijativna binarna operacija, $\id(x)=x$ $\forall x\in [0,1]$ je nevtralni element, za vsak element obstaja inverz.

\smallskip
(b) $\id\in T$ $\Rightarrow$ $T\ne\emptyset$; 

$\forall \alpha,\beta\in T$ $(\alpha\beta)(0)=0$; 

$\id(0)=0$ $\Rightarrow$ $(\alpha^{-1}\alpha)(0)=0$ $\Rightarrow$ $\alpha^{-1}(\alpha(0))=0$ $\Rightarrow$ $\alpha^{-1}(0)=0$ $\Rightarrow$ $\alpha^{-1}\in T$.


\smallskip
(c) Naj bo $f:[0,1]\Rightarrow[0,1]$ bijekcija t.d. $f(0)=1$. Potem $fT\subseteq\{\sigma\in S_{[0,1]}\,|\,\sigma(0)=1\}$ in $\{\sigma\in S_{[0,1]}\,|\,\sigma(0)=1\}\subseteq fT$...

\smallskip
(d) $[S_{[0,1]}:T]=\infty$.







\bigskip
\begin{zad}
Konstruiraj Cayley-evo tabelo za alternirajoči grupi $A_2$ in $A_3$. Ali je $\Z_3\cong A_3$? (Odgovor razloži.)
\end{zad}


\medskip
Re. 


\smallskip
$|A_2|=\frac{2!}{2}=1$, $A_2=\{(1)\}$;

$|A_3|=\frac{3!}{2}=3$, $A_3=\{(1),(123),(132)\}$;

\smallskip
\begin{tabular}{ c | c c c }			
  $\cdot$ & $(1)$\\
    \hline
   $(1)$ & $(1)$
\end{tabular}\qquad
\begin{tabular}{ c | c c c }			
  $\cdot$ & $(1)$ & $(123)$ & $(132)$\\
    \hline
   $(1)$ & $(1)$ & $(123)$ & $(132)$\\
   $(123)$ & $(123)$ & $(132)$ & $(1)$\\
   $(132)$ & $(132)$ & $(1)$ & $(123)$
\end{tabular}

\begin{tabular}{ c | c c c }			
  $+$ & $0$ & $1$ & $2$\\
    \hline
   $0$ & $0$ & $1$ & $2$\\
   $1$ & $1$ & $2$ & $0$\\
   $2$ & $2$ & $0$ & $1$
\end{tabular}
$\Rightarrow$ $\Z_3\cong A_3$.

\bigskip
\begin{zad}
Naj bo $\R^*$ grupa neničelnih realnih števil glede na operacijo množenja. Uporabi prvi izrek o izomorfizmu, in pokaži da je
$$
\R^*/\langle-1\rangle\cong\R^+.
$$
\end{zad}


\medskip
Re. $\langle -1 \rangle=\{1,-1\}$, $\{x\in\R^*\,:\,|x|=1\}=\{1,-1\}$.

Naj bo $\phi:\R^*\longrightarrow\R^+$ definiran s $x\longrightarrow |x|$. Potem je $\phi$ homomorfizem grup, $\ker(\phi)=\langle -1 \rangle$ in $\phi(\R^*)=\R^+$...





\newpage
\bigskip
\begin{zad}
Pokaži, da je vsaka abelska grupa reda 15 ciklična.
\end{zad}


\medskip
Re. 

\smallskip
$|G|=15=3\cdot 5$, 

Cauchijev izrek za abelske grupe $\Rightarrow$ $\exists a,b\in G$, $a\ne e$, $b\ne e$, $|a|=3$, $|b|=5$...

Naj bo $c=ab$. Potem $|c|\not\in\{1,3,5\}$...

$|c|=15$ $\Rightarrow$ $G=\langle c\rangle$.






\vfill{Več na
\url{http://osebje.famnit.upr.si/~penjic/}
}




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