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\newcommand{\ZZ}{\mathbb{Z}}
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\noindent
Math 541 \\
Problem Set 2
\begin{enumerate}


\item[0.3.12.] Since $a$ and $n$ are not relatively prime, there is an integer $d > 1$ such that $d \mid a$ and $d \mid n$.  Observe that $d \le n$.  Take $b = n/d$, so $1 \le b < n$.  Then $ab = a (n/d) = n (a/d) \equiv 0 \pmod n$.

Now suppose there is a $c$ such that $ac \equiv 1 \pmod n$.  Then mod $n$, we have
\[ b \equiv 1 \cdot b \equiv (ac)b \equiv (ab)c \equiv 0 \cdot c \equiv 0, \]
but this is not true since $1 \le b < n$.


\item[0.3.15.]
\begin{enumerate}
\item Since 13 is prime and $20 = 2^2 \cdot 5$, they have no common prime factors, hence are relatively prime.  17 is an inverse since $13 \cdot 17 = 221 \equiv 1 \pmod{20}$.
\end{enumerate}


\item[1.1.1.] 
\begin{enumerate}
\item No.  $(5 - 3) - 1 = 1$, but $5 - (3 - 1) = 3$.
\item Yes.
\begin{align*}
(a \star b) \star c &= (a + b + ab) \star c = a + b + ab + c + ac + bc + abc \\
a \star (b \star c) &= a \star (b + c + bc) = a + b + c + bc + ab + ac + abc
\end{align*}
which are equal.
\item No.  $(1 \star 1) \star 2 = 2/5 \star 2 = 12/25$, but $1 \star (1 \star 2) = 1 \star 3/5 = 8/25$.
\item Yes.  This is just addition of fractions.  For a proof,
\begin{align*}
[(a,b) \star (c,d)] \star (e,f) &= (ad + bc, bd) \star (e,f) = ((ad + bc)f + bde, bdf) \\
(a,b) \star [(c,d) \star (e,f)] &= (a,b) \star (cf + de, df) = (adf + b(cf + de), bdf)
\end{align*}
which are equal.
\item No.  $(8/4)/2 = 1$, but $8/(4/2) = 4$.
\end{enumerate}


\item[1.1.6.] Most of these sets are not closed under addition.
\begin{enumerate}
\item Yes.  Suppose that $b$ and $d$ are odd.  Then when we write $a/b + c/d = (ad+bc)/(bd)$ in lowest terms, the denominator will divide $bd$, hence will be odd.  Thus the set is closed under addition.  Addition in $\QQ$ is associative.  The additive identity $0$ is in the set by hypothesis.  If $a/b$ is in lowest terms with $b$ odd then its additive inverse $(-a)/b$ is still in lowest terms.
\item No.  $1/6 + 1/6 = 2/6 = 1/3$.
\item No.  $1/2 + 1/2 = 1$.
\item No.  $3/2 + (-1) = 1/2$.
\item Yes.  Every element of this set can be written uniquely as $n/2$, where $n \in \ZZ$.  The set is closed under addition because $\ZZ$ is: $m/2 + n/2 = (m+n)/2$.  Addition in $\QQ$ is associative.  The additive identity $0 = 0/2$ is in the set.  If $n/2$ is in the set then so is its additive inverse $(-n)/2$.
\item No.  $1/2 + 1/3 = 5/6$.
\end{enumerate}


\item[1.2.2.] An arbitrary element $x \in D_{2n}$ can be written in terms of the generators as $x = s^{k_1} r^{i_1} s^{k_2} r^{i_2} \dotsb s^{k_m} r^{i_m}$ where the powers $k_1, k_2, \dotsc, k_m, i_1, i_2, \dotsc, i_m \in \ZZ$.  Since $rs = sr^{-1}$, we can move all the $s$'s to the left and $r$'s to the right, so $x = s^k r^i$ with $k, i \in \ZZ$.  Since $s^2 = 1$, we have $x = r^i$ or $x = sr^i$.  If $x$ is not a power of $r$ then $rx = rsr^i = s r^{-1} r^i = s r^i r^{-1} = x r^{-1}$.


\end{enumerate}
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