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\def\Mat{{\mbox{\rm Mat}}}
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\begin{document}




\raggedright


\renewcommand{\thesection}{}
\section{\centerline{Algebra III - Abstraktna algebra, 16.02.2017.}}










\bigskip
\begin{zad}
\begin{itemize}
\item[{\rm (a)}] Pokaži, da je množica $\A=\{\left[\begin{matrix} x&y\\ 0&1 \end{matrix}\right]\,|\,x,y\in\R,\, x\ne 0\}$ zaprta za operacijo (običajnega) matričnega množenja. Ali je operacija matričnega množenja komutativna na množici $\A$? Ali ima vsak element iz množice $\A$ inverz? Odgovore utemelji!
\item[{\rm (b)}] Dana je polgrupa $(\{f:\R\rightarrow[0,1]\},\circ)$. Poišči leve in desne enote v tej polgrupi. (Spomnimo se: Za $(M,\circ)$ rečemo, da je polgrupa, če je operacija $\circ$ zaprta in asociativna.)
\end{itemize}
\end{zad}


\medskip
Re. 

\smallskip
(a) je zaprta, ni komutativna, inverz je $\left[\begin{matrix} \frac{1}{x}&-\frac{y}{x}\\0&1\end{matrix}\right]\in\A$.

\smallskip
(b) Leva identiteta je $e(x)=\left\{ \begin{array}{ll}
x & \hbox{če } \; x\in[0,1], \\
1 & \hbox{če } \; x\in\R\backslash[0,1].
\end{array} \right.$ 

Polgrupa nima desne identitete.


\bigskip
\begin{zad}
Naj bosta $A$ in $B$ končni ciklični grupi, za kateri velja $|A|=n$ in $|B|=m$. 
\begin{itemize}
\item[{\rm (a)}] Če je $n=2$ in $m=3$ pokaži, da je $A\times B$ ciklična grupa.
\item[{\rm (b)}] Če je $n=2$ in $m=2$ pokaži, da $A\times B$ ni ciklična grupa.
\item[{\rm (c)}] Ali je $A\times B$ ciklična, če je $n=4$ in $m=6$?
\end{itemize}
\end{zad}


\medskip
Re. lahko uporabimo izrek $|(a,b)|=\lcm(|a|,|b|)$.


\smallskip
(a) $|A|=2$ $\Rightarrow \exists a\in A$ t.d. $|a|=2$;
$|B|=2$ $\Rightarrow \exists b\in B$ t.d. $|b|=3$;... 
$A\times B=\langle(a,b)\rangle$.  

\smallskip
(b) $\forall (a,b)\in A\times B$ imamo $|(a,b)|\le 2$; ni ciklična


\smallskip
(c) $\forall (a,b)\in A\times B$ imamo $|(a,b)|\le 12$; ni ciklična




\bigskip
\begin{zad}
Naj bo dana grupa $G=\{z\in\C\,|\,|z|=1\}$ za operacijo množenja kompleksnih števil in podmnožica $H=\{1,-1\}\subseteq G$. Pokaži, da je $H$ podgrupa edinka v $G$ in da velja $G/H\cong G$.
\end{zad}


\medskip
Re. $zHz^{-1}=z\{1,-1\}z^{-1}=H$; Definirajmo $\phi:G\rightarrow G$ t.d. $\phi(z)=z^2$. Potem za $\forall x\in G$ $\phi(x)\in G$, $\phi$ je homomorfizem, $\phi(G)=G$ in $\ker(\phi)=H$. Zdaj uporabimo prvi izrek o izomorfizmu.





%%\newpage
\bigskip
\begin{zad}
Pokaži, da mora grupa reda $63$ vsebovati element reda $3$.
\begin{itemize}
\item[{\rm (a)}] Z uporabo Cauchy-evega izreka.
\item[{\rm (b)}] Brez uporabe Cauchy-evega izreka ali izreka Sylowa.
\end{itemize}
\end{zad}


\medskip
Re. 

\smallskip
(b) Naj bo $G$ dana grupa, $|G|=63$. Lagrangeov izrek $\Rightarrow$ če je $a\in G$ potem $|a|\in\{1,3,7,9,21,63\}$. Prvo moramo pokazati, da $G\backslash\{e\}$ ima najmanjše en element, ki ni reda 7...






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




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