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\begin{document}
\title{Exam 2 solutions}
\author{Math 116: Finashin, Pamuk, Pierce, Solak}
\date{Tuesday, May 3, 2011}
\maketitle

\textbf{Instructions:}
Please, be accurate and show clearly the logic of your solutions.
Only the answers are not enough: indicate your calculations and
arguments.

\begin{problem}{\bf(9 pts)}
  Of the set $\{1,2,3,4,5,6,7,9\}$, let $\sigma\in S_9$ be the permutation
  \begin{equation*}
    \begin{bmatrix}
      1&2&3&4&5&6&7&8&9\\
      3&7&4&1&6&8&2&9&5
    \end{bmatrix}.
  \end{equation*}
  \begin{enumerate}
  \item
Write $\sigma$ as a product of disjoint cycles.
%\vskip25mm
\begin{solution}

$\sigma=(1 3 4)(2 7)(5 6 8 9)$.

\end{solution}
%\item
%Write $\sigma$ as a product of transpositions.
%\item Find the order of $\sigma$.\vskip20mm
%What can you conclude about the
%index of the cyclic subgroup generated by $\sigma$,
%$<\sigma>\subset S_9$ ?
 \item Find $\sigma^{100}$. (Hint: find the order of $\sigma$).
 %\vskip35mm
 \begin{solution}
 The length of $\sigma$ is the least common multiple of the length of
 the disjoint cycles. Hence $|\sigma|= \lcm (3,2,4)=12$. Thus
 $$\sigma^{100}= \sigma^{96}\sigma^{4}=\sigma^{4}=(1 3 4).$$
\end{solution}
 \item
Determine whether permutation $\sigma$ is even or odd.
 \begin{solution}
Since we can write $\sigma= (1 4) (1 3)  (2 7) (5 9) (5 8) (5 6)$ as
a product of even number of transpositions, $\sigma$ is an even
permutation.
\end{solution}
%\vskip25mm
  \end{enumerate}
\end{problem}

\begin{problem}{\bf(6 pts)}
Consider transpositions $\alpha=(12)$, $\beta=(23)$, $\gamma=(34)$
in $S_4$.
\begin{enumerate}
 \item Find the product $\alpha\beta\gamma$.
 %\vskip30mm
 \begin{solution}
$\alpha\beta\gamma=(12)(23)(34)=(1234)$.
\end{solution}
 \item
 Find $g\in S_4$ such that $\beta=g\alpha g^{-1}$.
 \begin{solution}
Note that $g$ is not unique. Since $g(1)=2$ and $g(2)=3$  we may
write $g$ as  $$g=(12)(23)=(123)$$.
\end{solution}
\end{enumerate}
\end{problem}

\pagebreak


\begin{problem}{\bf(8 pts)}
\begin{enumerate}
 \item
 Find the number of cosets for subgroups $3\Z\subset\Z$ and
 $\Z\subset\Q$. In the first case, give an explicit list of these
 cosets.
%\vskip45mm
\begin{solution}
The cosets of $3\Z$ in $\Z$ are $3\Z$, $1+3\Z$, and $2+3\Z$: there are
three of them.  

$\Z$ has infinitely many cosets in $\Q$ [since if $a$ and $b$ are distinct
  elements of $\Q$ between $0$ and $1$, then $a+\Z$ and $b+\Z$ are
  distinct cosets].
\end{solution} 
\item
Find the group multiplication table of $\Z/3\Z$ (pay attention
that the group operation is addition).
 \begin{solution}
$
   \begin{array}[t]{r|rrr}
     &  3\Z&1+3\Z&2+3\Z\\\hline
3\Z  &  3\Z&1+3\Z&2+3\Z\\
1+3\Z&1+3\Z&2+3\Z&  3\Z\\
2+3\Z&2+3\Z&  3\Z&1+3\Z
   \end{array}$
\end{solution}
 \item Prove that every element of $\Q/\Z$ has finite
 order.
% \vskip45mm
 \begin{solution}
Every element of $\Q/\Z$ is $k/n+\Z$ for some $k$ and $n$ in $\Z$ such
that $n>0$.  Then $n(k/n+\Z)=k+\Z=\Z$, the identity in $\Q/Z$.  Thus
the order of $k/n+\Z$ is less than or equal to $n$: in particular, it
is finite.
\end{solution}
\end{enumerate}
\end{problem}

\begin{problem}{\bf(7 pts)}
Assume that $G$ is a group of prime order $p$.
\begin{enumerate}
 \item
What can be the order of an element $g\in G$ ?
(Explain~!)
%\vskip40mm
\begin{solution}
By the Lagrange Theorem, the order of an element divides the order of
a group.  In particular, the order of an element of $G$ divides $p$;
so the order is $0$ or $1$.
\end{solution}
 \item
 Prove that $G$ is a cyclic group.
 \begin{solution}
The only element of $G$ of order $1$ is the identity, $e$.  If $a\in
G\smallsetminus\{e\}$, then the order of $a$ is $p$, by part (a).  So the cyclic
subgroup $\langle a\rangle$ of $G$ has the same order as $G$ itself;
so that cyclic subgroup \emph{is} $G$.
\end{solution}
 %\vskip50mm
\end{enumerate}
\end{problem}

\pagebreak


\begin{problem}{\bf(10 pts)}
Suppose that $H$ and $N$ are subgroups of a group $G$.
 Assume that subgroup $N$ is normal, and let $K=H\cap N$.
  \begin{enumerate}
 \item
 Prove that $K$ is a normal subgroup of $H$.
% \vskip45mm
 \begin{solution}
Since $H,N$ are subgroups of $G$ then $e\in H$ and $e\in N$, so $e\in H\cap N=K$ i.e. $K$ is non-empty.
Let $a,b\in K$ consider $ab^{-1}$, since $a,b\in H$ and $a,b\in N$ and they are subgroups of $G$ then $ab^{-1}\in H$ and $ab^{-1}\in N$, so $ab^{-1}\in K$. Therefore, $K$ is a subgroup of $H$.\\
Let $h\in H$ and $a\in K$, then $a\in H$ and $a\in N$, so $hah^{-1}\in H$ and since $N$ is normal $hah^{-1}=n$ for some $n\in N$ which means $hah^{-1}\in N$, therefore $hah^{-1}\in H\cap N=K$. Hence,  $K$ is a normal subgroup of $H$.(OR: $ha=bh$ for some $b\in N$ since $N$ is normal, but $b=hah^{-1}$ which is in $H$ as well, so $b\in K$, hence $hK=Kh$ for all $h\in H$.)
\end{solution}
 \item
Let $A=\{x\in G\colon xK=Kx\}$. Prove that $H\included
A$.
\begin{solution}
  By part (a)$K$ is normal in $H$, so we know that $hK=Kh$ for all $h\in H$, but then $h\in A$ for all $h\in H$. Hence $H\included A$.
\end{solution}
%\vskip45mm
 \item
Prove that $A$ is a subgroup of $G$.
%\vskip45mm
\begin{solution}
  \begin{enumerate}
  \item
Since $eK=K=Ke$, we have $e\in A$.
\item
If $x\in A$, so that $xK=Kx$, then
$K=x^{-1}xK=x^{-1}Kx$, so $Kx^{-1}=x^{-1}Kxx^{-1}=x^{-1}K$, and so
$x^{-1}\in A$.
\item
If $x,y\in A$, then $xK=Kx$ and $yK=Ky$, so $xyK=xKy=Kxy$, hence $xy\in A$.
  \end{enumerate}

\end{solution}
  \end{enumerate}
\end{problem}


\begin{problem}{\bf(5 pts)}
Let $H$ and $K$ be normal subgroups of $G$ such that $G/H$ has
order 5, and $G/K$ has order 3. Prove that for any $g\in G$,
$g^{15}\in K\cap H$.
\begin{solution}
Let $g\in G$, consider the element $gH\in G/H$. Since $G/H$ has
order 5 then by Lagrange Thm. $o(gH)|5$ by problem 4(a)  $o(gH)=5$ or  $o(gH)=1$.\\
If $o(gH)=1$ then $gH=eH=H\Rightarrow g\in H\Rightarrow g^5\in H$. If $o(gH)=5$ then since $H$ is a normal subgroup of $G$
$$eH=(gH)^5=gHgHgHgHgH=gHgHgHggH=\cdots=g^5H$$ which means $g^5\in H$.\\

Similarly, $o(gK)|3$ then $o(gK)=3$ or  $o(gK)=1$. If $o(gK)=1$ then $g^3\in K$.  If $o(gK)=3$ then since $K$ is a normal subgroup of $G$ $eK=(gK)^3=g^3K$ which means $g^3\in K$.\\

Now, since $g^15=(g^5)^3\in H$ and $g^15=(g^3)^5\in K$ then $g^15\in
K\cap H$. 
\end{solution}
\end{problem}

\pagebreak

\begin{problem}{\bf(6 pts)} Consider subsets $4\Z\subset\Z$, $\{0\}\subset\Z$,
$\varnothing\subset\Z$, $\Z\subset\Q$, $(0,\infty)\subset\R$,
 $\{[0],[1],[2],[4]\}\subset\Z_6$,
$M_2(\Z)\subset M_2(\R)$.
\begin{enumerate}
 \item
Which of the above examples are NOT subrings? Why they are not
subrings~?
%\vskip40mm
\begin{solution}
$\varnothing\subset\Z$, because a subring should be non-empty,

$(0,\infty)\subset\R$, because for number $x\in(0,\infty)$ its
opposite, $-x$, does not belong to $(0,\infty)$,

$\{[0],[1],[2],[4]\}\subset\Z_6$, because
$[1]+[2]=[3]\notin\{[0],[1],[2],[4]\}$.
\end{solution}
 \item
Which ones are NOT ideals? Why not~?
%\vskip40mm
 \begin{solution}
An ideal is a subring, thus, the three examples in part (a) which
are not rings are also not ideals. There are two more examples:

 $\Z\subset\Q$ (the product of a rational and an integers may be NOT
 an integer), and

 $M_2(\Z)\subset M_2(\R)$ (the product of an integer matrix with a
 real matrix may be NOT an integer matrix).

(So, the only examples of ideals here are $4\Z\subset\Z$,
 and $\{0\}\subset\Z$.)
\end{solution}
\end{enumerate}
\end{problem}



\begin{problem}{\bf(9 pts)}
In each of the following two subrings of $\Z_{12}$, find a unity
or show that there is no unity (multiplicative identity).
\begin{enumerate}
 \item
 $2\Z_{12}=\{[0],[2],[4],[6],[8],[10]\}$
 %\vskip40mm
 \begin{solution} If $[2n]$ is the unity, then $[2][2n]=[2]$, that
 is $4n=2 \mod 12$. But $4n-2$ is not divisible by 4, and thus, by
 12. So, there is no unity in $2\Z_{12}$.
\end{solution}
 \item
 $3\Z_{12}=\{[0],[3],[6],[9]\}$
 \begin{solution} $[9]$ is the unity in $2\Z_{12}$,
 because $[9][0]=[0],
 [9][3]=[27]=[3], [9][6]=[54]=[6], [9][9]=[81]=[9]$.
\end{solution}
 %\vskip35mm
 \item Find all the zero divisors in the ring $\Z_{12}$.
  \begin{solution}
  The zero divisors are $[2],[3],[4],[6],[8],[9],[10]$, because in
  $\Z_{12}$, we have

  $[2][6]=[3][4]=[8][9]=[10][6]=[0]$.
\end{solution}
\end{enumerate}\end{problem}


\end{document}
