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{\bf Math 307 Abstract Algebra \quad Homework 7 \hfill
Name: Sample Texfile}
\bigskip
Five points for each question.
\begin{enumerate}
\item If $r$ is a divisor of $m$ and $s$ is a divisor of $n$, find a subgroup of
$\Z_m \oplus \Z_n$ that is isomorphic to $\Z_r \oplus \Z_s$.
[Example: In $\Z_6 \oplus \Z_6$, the subgroup $\{(2p,3q): p, q \in \Z\}$
is isomorphic to $\Z_3 \oplus \Z_2$ under the isomorphism
$\phi(2p,3q) = (p,q) \in \Z_3 \oplus \Z_2$.]
\item (a)
What is the order of the element $14+ \langle 8 \rangle$ in $\Z_{24}/ \langle 8 \rangle$?
(b) What is the order of $4U_5(105)$ in the factor group
$U(105)/U_5(105)$.
[Hint: The order of $aH$ is the smallest $m\in \N$
such that $a^m \in H$.]
\item Let $G = \Z_4 \oplus U(4)$, $H = \langle (2,3)\rangle$ and
$K = \langle (2,1) \rangle$. Show that $G/H \not\cong G/K$.
[Hint: $H = \{(2,3), (0,1)\}$ and
$K = \{(2,1), (0,1)\}$.
Show that $G/H$ is cyclic but $G/K$ is not.]
\item Let $G$ be a finite group, and $H$ be a normal subgroup of $G$.
(a) Show that the order of $aH$ in $G/H$ must divide the order of $a$ in $G$.
(b) Show that it is possible that $aH = bH$, but $|a| \ne |b|$.
[Hint: (a) Note that if $a^n = e$, then $(aH)^n = eH = H$.]
(b) Note that if $a \in H$ then $aH = eH$.]
\item If $G$ is a group and $|G:Z(G)| = 4$, prove that
$G/Z(G)$ is isomorphic to $\Z_2 \oplus \Z_2$.
[Hint: Note that a four element group is isomorphic to $\Z_4$ or $\Z_2\oplus \Z_2$.]
\item Suppose that $N \triangleleft G$ and $|G/N| = m$,
show that $x^m \in N$ for all $x \in G$.
\item (a) Let $G = \{3^a 6^b 10^c: a, b , c \in \Z\}$ under multiplication.
Show that $G$ is isomorphic to
\quad
$\langle 3 \rangle \times \langle 6 \rangle \times \langle 10 \rangle$.
(b) Let $H = \{3^a 6^b 12^c: a, b , c \in \Z\}$ under multiplication.
Show that $H$ is NOT isomorphic to
\quad $\langle 3 \rangle \times \langle 6 \rangle \times \langle 12 \rangle$.
[Hint: Check internal direct product
conditions.]
\item (Extra credits) (a) (5 points) Show that
$U(2^3) = \langle 7 \rangle \oplus \langle 3 \rangle$,
$U(2^4) = \langle 15 \rangle \oplus \langle 3 \rangle$.
[Note: $\langle 7 \rangle, \langle 3 \rangle$ are subgroups in $U(2^3)$
generated by $7$ and $3$ in the first part, and
$\langle 15 \rangle, \langle 3 \rangle$ are subgroups in $U(2^4)$
generated by $7$ and $3$ in the second part.]
(b) (5 points) Show that
$U(2^m) = \langle 2^m -1\rangle \oplus \langle 3 \rangle$.
\item (Extra credits) Let $n = p_1^{r_1} \cdots p_k^{r_k}$ such that
$p_1, \dots, p_k$ are distinct primes and $r_1, \dots, r_k$ are positive integer.
Determine $n$ such that $U(n)$ is a cyclic group. [Give explanation.]
\item (Extra credits)
Suppose $s,t$ are relatively prime. Prove that $U_t(st)$ is isomorphic to $U(s)$.
[Hint: $U_t(st) = \{ kt+1 \in U({st}): k \in \{0, \dots, s-1\}\}$.
Note that some of the $k$ should would not give $kt + 1 \in U(st)$.
Define $f: U_t(st) \rightarrow U(s)$ by $f(x) = [x]_s = y$, i.e.,
$x = sq + y$ with $y \in \{0, \dots, s-1\}$. We need to show that
$f$ is an isomorphism. One important step is to show that
$\gcd(x,st) = 1$ will ensure that $\gcd(y,s) = 1$ so that
$y \in U(s)$.
To show that $f$ is surjective, for any $y \in U(s)$, we need to find
$x = pt + 1 \in U_t(st)$ such that $f(x) = y$. Thus, we need to find
$[y]_s = [pt+1]_s = [pt]_s + [1]_s$. Note that $[y]_s \in U(s)$
has an inverse in $U(s)$. Also, $[t]_s \in U(s)$ also has an inverse in $U(s)$.
With these two pieces of information, one should be able to find $k$ such that
$x = ks + 1 \in U(st)$ satisfying $[x]_s = [y]_s \in U(s)$.
]
\end{enumerate}
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