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\begin{document}




\raggedright


\renewcommand{\thesection}{}
\section{\centerline{Algebra III - Abstraktna algebra, 29.08.2017.}}










\bigskip
\begin{zad}
\begin{itemize}
\item[{\rm (a)}] Naj bo $S$ množica moči $n\in\N$. Koliko različnih binarnih operacij lahko definiramo na množici $S$?
\item[{\rm (b)}] Dokažite, da tvorijo vse matrike oblike
$$\left[
\begin{matrix} 
\cos\alpha & -\sin\alpha\\
\sin\alpha & \cos\alpha
\end{matrix}
\right],\qquad \alpha\in\R
$$
grupo za običajno operacijo množenja matrik.
\end{itemize}
\end{zad}


\medskip
Re. 

\smallskip
(a) $|S|=n$ $\Rightarrow$ $|S\times S|=n^2$. Vsak od $n^2$ elementov se lahko preslika v $n$ različnih elementov; $n\cdot n\cdot ...\cdot n=n^{n^2}$ $\Rightarrow$ obstaja $n^{n^2}$ različnih binarnih operacij.

\smallskip
(b) $\left[
\begin{matrix} 
\cos\alpha & -\sin\alpha\\
\sin\alpha & \cos\alpha
\end{matrix}
\right]\cdot \left[
\begin{matrix} 
\cos\beta & -\sin\beta\\
\sin\beta & \cos\beta
\end{matrix}
\right]=\left[
\begin{matrix} 
\cos(\alpha+\beta) & -\sin(\alpha+\beta)\\
\sin(\alpha+\beta) & \cos(\alpha+\beta)
\end{matrix}
\right]$;

$
\left[
\begin{matrix} 
\cos0 & -\sin0\\
\sin0 & \cos0
\end{matrix}
\right]=
\left[
\begin{matrix} 
1 & 0\\
0 & 1
\end{matrix}
\right];
$

$
\left[
\begin{matrix} 
\cos\alpha & -\sin\alpha\\
\sin\alpha & \cos\alpha
\end{matrix}
\right]\cdot
\left[
\begin{matrix} 
\cos(-\alpha) & -\sin(-\alpha)\\
\sin(-\alpha) & \cos(-\alpha)
\end{matrix}
\right]=
\left[
\begin{matrix} 
1 & 0\\
0 & 1
\end{matrix}
\right]=
\left[
\begin{matrix} 
\cos(-\alpha) & -\sin(-\alpha)\\
\sin(-\alpha) & \cos(-\alpha)
\end{matrix}
\right]\cdot
\left[
\begin{matrix} 
\cos\alpha & -\sin\alpha\\
\sin\alpha & \cos\alpha
\end{matrix}
\right].
$








\bigskip
\begin{zad}
Podane so naslednje permutacije:
$$
\alpha=\left(
\begin{matrix} 
1&2&3&4\\
2&3&4&1
\end{matrix}
\right),\,\,\,
\beta=\left(
\begin{matrix} 
1&2&3&4\\
1&3&2&4
\end{matrix}
\right),\,\,\,
\gamma=\left(
\begin{matrix} 
1&2&3&4\\
2&1&4&3
\end{matrix}
\right),\,\,\,
\delta=\left(
\begin{matrix} 
1&2&3&4\\
3&1&2&4
\end{matrix}
\right).
$$
\begin{itemize}
\item[{\rm (a)}] Zapišite ciklične dekompozicije permutacij.
\item[{\rm (b)}] Permutacije izrazite kot produkt transpozicij.
\item[{\rm (c)}] Izračunajte $\alpha\beta$, $\beta\alpha$, $\alpha^{-1}\beta$, $\beta\alpha^2$, $(\alpha\beta)^{-2}$, $\alpha\beta\alpha^3$, $\gamma^{-1}\beta\gamma^{-1}$ in $\gamma\beta^2\alpha$.
\end{itemize}
\end{zad}


\medskip
Re. 

\smallskip
$\alpha=(1234)$, $\beta=(23)$, $\gamma=(12)(34)$, $\delta=(132)$.

\smallskip
$\alpha=(14)(13)(12)$, $\beta=(23)$, $\gamma=(12)(34)$, $\delta=(12)(13)$.

\smallskip
$\alpha\beta=(124)$, $\beta\alpha=(134)$, $\alpha^{-1}\beta=(143)$, $\beta\alpha^2=(1243)$, $(\alpha\beta)^{-2}=(124)$, $\alpha\beta\alpha^3=(34)$, $\gamma^{-1}\beta\gamma^{-1}=(14)$ in $\gamma\beta^2\alpha=(24)$.


\bigskip
\begin{zad}
\begin{itemize}
\item[{\rm (a)}] Poišči odseke podgrupe $\langle4\rangle$ v grupi $\Z_{12}$ in konstruiraj Cayley-evo tabelo za kvocientno grupo $\Z_{12}/\langle4\rangle$.
\item[{\rm (b)}] Če obstaja, poišči kak netrivialni homomorfizem $\varphi: \Z_{12}\rightarrow \Z_4$.
\end{itemize}
\end{zad}


\medskip
Re. 

\smallskip
(a) $\{0,4,8\},\{1,5,9\},\{2,6,10\},\{3,7,11\}$,


\begin{tabular}{ c | c c c c}			
  $+$ & $\langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$\\
    \hline
  $\langle 4 \rangle$ & $\langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$\\
  $1+\langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$ & $\langle 4 \rangle$\\
  $2+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$ & $\langle 4 \rangle$ & $1+\langle 4 \rangle$\\
  $3+\langle 4 \rangle$ & $3+\langle 4 \rangle$ & $\langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$
\end{tabular}



\medskip
(b) Naj bo $\varphi(1)=a$. 

$12=1+1+...+1=0$, $\varphi(0)=\varphi(12)=\varphi(1+...+1)=\varphi(1)+...+\varphi(1)=12a=0$. Če je $a=1$, potem $\varphi(j)=j$ je homomorfizem.



\bigskip
\begin{zad}
Diederska grupa $G = D_6$ na naraven način deluje na množici oglišč pravilnega 6-kotnika, označimo jo z $X=\{1,2,3,4,5,6\}$. Enako velja za njene podgrupe $H = \langle\rho\rangle$, $K = \langle\tau\rangle$ in $L = \langle\tau\rho\rangle$ kjer je $\rho = (1 2 3 4 5 6)$ in $\tau = (2 6)(3 5)$.
\begin{itemize}
\item[{\rm (a)}] Poišči orbite delovanj grup $G$, $H$, $K$ in $L$ na $X$!
\item[{\rm (b)}] Spomnimo se: Delovanje je tranzitivno, če za poljubna $x_1,x_2\in X$, obstaja $g\in G$, da je $x_1=g*x_2$. Katera izmed delovanj $G$, $H$, $K$ in $L$ so tranzitivna?
\item[{\rm (c)}] Spomnimo se: Delovanje je zvesto, če $\forall g\in G :$ $g*x=x$ $\forall x\in X$ $\Rightarrow$ $g=e$. Je delovanje $G$ na $X$ zvesto? Kaj pa ostala delovanja?
\end{itemize}
\end{zad}


\medskip
Re. 

\smallskip
(a) Orbite delovanj grup $G$ je $X$; orbite delovanj grup $H$ je $X$; orbite delovanj grup $K$ so $\{1\}$, $\{2,6\}$, $\{3,5\}$ in $\{4\}$; orbite delovanj grup $L$ so $\{1,6\}$, $\{2,5\}$ in $\{3,4\}$. 


\smallskip
(b) H in K sta tranzitivna, K in L nista.

\smallskip
(c) Vse štiri grupe so zvesta.


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




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