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\begin{document}

\title{Cebir I (MAT 531)}
\date{27 Aral\i k 2013}
\author{David Pierce}
\maketitle
\thispagestyle{empty}

\begin{problem}
Prove or disprove:
\begin{enumerate}
\item
$(\Qp,\cdot)$ is a free abelian group.
\item
$(\Q,+)$ is a free abelian group.
\item
There are exactly $9$ nonisomorphic abelian groups of order $1000$.
\item
All groups of order $143$ are abelian.
\item
All groups of order $143$ are cyclic.
\item
There is a simple group of order $385$.
\end{enumerate}
\end{problem}

\begin{solution}\mbox{}
\begin{enumerate}
\item
$(\Qp,\cdot)$ is a free abelian group, since if $(p_i\colon i\in\upomega)$ is the list of primes, the map
\begin{equation*}
(x_i\colon i\in\upomega)\mapsto\prod_{i\in\upomega}p_i{}^{x_i}
\end{equation*}
from $\sum_{i\in\upomega}\Z$ to $\Qp$ is an isomorphism.
\item
$(\Q,+)$ is not a free abelian group.
Indeed, suppose $h$ is an embedding of a free abelian group $\Z\oplus\Z\oplus\cdots$ in $\Q$.
If $h(1,0,\dots)=a$, then $2b=a$ for some $b$ in $\Q$,
but $b$ is not in the image of $h$.
\item
There are exactly $9$ nonisomorphic abelian groups of order $1000$.
For, $1000=2^3\cdot 5^3$, so (by the Chinese Remainder Theorem) 
an abelian group of order $1000$ 
is the direct sum of an abelian group of order $2^3$ 
and an abelian group of order $5^3$.
there are three possibilities for each of these
(by the classification of finitely generated abelian groups),
namely
\begin{align*}
&\Z_{p^3},&&\Z_p\oplus\Z_{p^2},&&\Z_p\oplus\Z_p\oplus\Z_p.
\end{align*}
\item
All groups of order $143$ are abelian.
For, $143=11\cdot13$, and $13\not\equiv1\pmod{11}$, 
so the Sylow subgroups are normal subgroups, 
and the group is isomorphic to $\Z_{11}\oplus\Z_{13}$.
\item
All groups of order $143$ are cyclic,
since as we have just shown, they are isomorphic to $\Z_{11}\oplus\Z_{13}$,
which is isomorphic to $\Z_{143}$ by the Chinese Remainder Theorem.
\item
There is no simple group of order $385$.
For, $385=5\cdot7\cdot11$.
Let $n$ be the number of Sylow $7$-subgroups.
Then
\begin{align*}
  n&\divides 55,&n&\equiv1\pmod 7.
\end{align*}
From the first condition, $n\in\{1,5,11,55\}$.  Therefore $n=1$.
Thus there is a unique Sylow $7$-subgroup,
which is a proper normal subgroup.
\end{enumerate}
\end{solution}
\pagebreak
\begin{problem}
Describe the following groups as products of cyclic groups.
\begin{enumerate}
\item
$\gpres{a,b}{a^7,b^5,a^4ba^6b^4}$
\item
$\gpres{a,b}{a^{11},b^5,a^4ba^{10}b^4}$
\end{enumerate}
\end{problem}

\begin{solution}\mbox{}
\begin{enumerate}
\item
Let $G=\gpres{a,b}{a^7,b^5,a^4ba^6b^4}$.
Here $a^4=bab\inv$.  
Thus all elements of $G$ are of the form $a^ib^j$, and moreover
\begin{equation*}
  G=\gpgen a\rtimes\gpgen b.
\end{equation*}
Here $\gpgen a$ is either isomorphic to $\Z_7$ or trivial,
and $\gpgen b$ is either isomorphic to $\Z_5$ or trivial.
Since the map $x\mapsto 4x$ is an automorphism of $\Z_7$ of order $3$,
and $3\ndivides5$, $\gpgen a$ will be trivial.
Indeed, using $ba^kb\inv=a^{4^k}$, we have
\begin{align*}
a&=b^3ab^{-3},&a^4&=b^4ab^{-4},&a^2&=a^{16}=b^5ab^{-5}=a.
\end{align*}
Hence $a=\gid$, so $G=\gpres b{b^5}$, which is isomorphic to $\Z_5$.
\item
Now let 
$G=\gpres{a,b}{a^{11},b^5,a^4ba^{10}b^4}$.
Here again $a^4=bab\inv$, but now $x\mapsto 4x$ is an automorphism of $\Z_{11}$ of order $5$.  So $G\cong\Z_{11}\rtimes\Z_5$.
(Strictly we use van Dyck's Theorem here,
according to which
there is an epimorphism from $G$ to $\Z_{11}\rtimes\Z_5$
taking $a$ to $(1,0)$ and $b$ to $(0,1)$.
Since $G$ has at most as many elements as  $\Z_{11}\rtimes\Z_5$,
the epimorphism must be an isomorphism.)
\end{enumerate}
\end{solution}



\end{document}