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\begin{document}




\raggedright


\renewcommand{\thesection}{}
\section{\centerline{Algebra III - Abstraktna algebra, 12.01.2018.}}








\begin{zad}
Poišči indeks podgrupe $\langle 3 \rangle$ v grupi $\Z_{24}$. Poišči indeks podgrupe $\langle (2,3) \rangle$ v grupi $\Z_{4}\times\Z_{6}$. Napiši Cayley-evo tabelo za kvocientno grupo $\Z_{24}/\langle 3 \rangle$.
\end{zad}






\medskip
Re. 

\smallskip
$\langle 3\rangle=\{0,3,6,9,12,15,18,21\}$, $|\langle 3\rangle|=8$, $[\Z_{24}:\langle 3\rangle]=3$.

\smallskip
$\langle (2,3)\rangle=\{(0,0),(2,3)\}$, $|\langle (2,3)\rangle|=2$, $[\Z_{4}\times\Z_6:\langle (2,3)\rangle]=12$.

\medskip
{\footnotesize\begin{tabular}{c|ccc|}
+ & $\langle 3\rangle$ & $1+\langle 3\rangle$ & $2+\langle 3\rangle$ \\\midrule
$\langle 3\rangle$ & $\langle 3\rangle$ & $1+\langle 3\rangle$ & $2+\langle 3\rangle$ \\ 
$1+\langle 3\rangle$ & $1+\langle 3\rangle$ & $2+\langle 3\rangle$ & $\langle 3\rangle$ \\ 
$2+\langle 3\rangle$ & $2+\langle 3\rangle$ & $\langle 3\rangle$ & $1+\langle 3\rangle$ \\\midrule
\end{tabular}}






\begin{zad}
\begin{itemize}
\item[{\rm (a)}] Naj bo $G$ grupa reda $77$ in naj bo $H$ podgrupa grupe $G$ reda $11$. Pokaži, da je $H$ edinka grupe $G$.

(Spomnimo se: Za končne podgrupe $H$ in $K$ grupe $G$ velja $$|HK|=\frac{|H||K|}{|H\cap K|}$$ kjer je $HK=\{hk\,|\,h\in H,\,k\in K\}.$)
\item[{\rm (b)}] Naj bosta $H$ in $K$ edinki grupe $G$, tako da velja $H\cap K=\{e\}$, $|H|=p$ ($p$ je praštevilo) in $|K|=4$. Če je $HK=G$, potem pokaži, da je $G$ abelska.

(Spomnimo se: Če sta  $H$ in $K$ edinki grupe $G$ in če je $H\cap K=\{e\}$, potem je $hk=kh$ za vsak $h\in H$ in $k\in K$).
\end{itemize}
\end{zad}




\medskip
Re. 

\smallskip
(a) $11$ deli $77$, $\exists a$ t.d. $|a|=11$, $|\langle a\rangle|=11$, $H:=\langle a\rangle$.

\smallskip
Če je $|K|=11$ in $H\ne K$ potem $|H\cap K|=1$, $7=|G|\ge |HK|=121$, protislovje.

\smallskip
Obstaja natanko ena grupa reda $11$. $\forall g\in G$, $gHg^{-1}$ je podgrupa reda 11,... $\Rightarrow$ $H\lhd G$.

\bigskip
(b) $|H|=p$, p je praštevilo $\Rightarrow$ $H$ je abelska.

\smallskip
$|K|=4$ $\Rightarrow$ bodisi $K\cong \Z_4$ ali $K\cong \Z_2\times\Z_2$ $\Rightarrow$ $K$ je abelska.

\smallskip
$g_1g_2=(h_1k_1)(h_2k_2)=...=(h_2h_1)(k_1k_2)=...=(h_2k_2)(h_1k_1)=g_2g_1$ \qquad$\Rightarrow$\qquad $G$ je abelska.





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[e]{{
\begin{zad}
Naj bo $\O$ grupa vseh simetrij tetraedra (rotacija, zrcaljenje, drsno zrcaljenje,...). Grupa $\O$ deluje na množici $\{v_1,v_2,v_3,v_4\}$ ogljišč tetraedra. Določi stabilizator ogljišča $v_1$ glede na to delovanje. Uporabi orbita-stabilizator izrek in izračunaj $|\O|$.
\end{zad}
}}
{fig:text}



\medskip
Re. 

\smallskip
$|\O|=|\O1|\cdot |\O_1|=4|\O_1|$, \qquad
$|\O_1|=|\O_12|\cdot |\O_{12}|=3|\O_{12}|$, \qquad
$|\O_{12}|=|\O_{12}3|\cdot \O_{123}=2\cdot 1$ 

$\Rightarrow$ $|\O|=24$ $\Rightarrow$ $|\O_1|=6$,

\smallskip
$\O_1=\left\{
\left(\begin{matrix} 1&2&3&4\\1&2&3&4\end{matrix}\right),
\left(\begin{matrix} 1&2&3&4\\1&2&4&3\end{matrix}\right), 
\left(\begin{matrix} 1&2&3&4\\1&3&2&4\end{matrix}\right), 
\left(\begin{matrix} 1&2&3&4\\1&3&4&2\end{matrix}\right), 
\left(\begin{matrix} 1&2&3&4\\1&4&2&3\end{matrix}\right), 
\left(\begin{matrix} 1&2&3&4\\1&4&3&2\end{matrix}\right)
\right\}$.








\begin{zad}
Naj bo $G$ enostavna grupa reda $168$. Koliko elemntov reda $7$ je v $G$?
\end{zad}




\medskip
Re. 

\smallskip
$|G|=2^3\cdot 3\cdot 7$, $|G|=7\cdot 24$, $\gcd(7,24)=1$ $\Rightarrow$ obstaja najmanj ena podgrupa $H$ reda $7$, $H$ je Sylowa $7$ podgrupa.

\smallskip
$|G|=7\cdot 24$, $\gcd(7,24)=1$ $\Rightarrow$ število Sylowih 7-podgrup je oblike $1+7m$, $1+7m$ deli $168$ $\Rightarrow$ ... število Sylowih $7$-podgrup je $8$ $\Rightarrow$ obstaja $7+7\cdot 6=49$ elementov reda $7$.




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




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