Dummit And Foote Solutions Chapter 4 Overleaf High Quality [BEST]

\subsection*Exercise 4.4.7 \textitShow that $\Aut(\Z/8\Z) \cong \Z/2\Z \times \Z/2\Z$.

\beginsolution Let $G = \langle g \rangle$ be a cyclic group. Then every element $a, b \in G$ can be written as $a = g^m$, $b = g^n$ for some integers $m, n$. Then \[ ab = g^m g^n = g^m+n = g^n+m = g^n g^m = ba. \] Thus $G$ is abelian. \endsolution

\subsection*Exercise 4.5.9 \textitLet $G$ be a finite group and let $H$ be a subgroup of $G$ with $

Subgroup lattice (inclusion): \[ \beginarrayc \Z_12 \\ \vert \\ \langle 2 \rangle \\ \vert \\ \langle 3 \rangle \quad \langle 4 \rangle \\ \vert \quad \vert \\ \langle 6 \rangle \\ \vert \\ \0\ \endarray \] Note: $\langle 3 \rangle$ contains $\langle 6 \rangle$ and $\langle 4 \rangle$ also contains $\langle 6 \rangle$. \endsolution Dummit And Foote Solutions Chapter 4 Overleaf High Quality

\subsection*Exercise 4.1.3 \textitFind all subgroups of $\Z_12$ and draw the subgroup lattice.

\subsection*Exercise 4.2.6 \textitLet $G$ be a group and let $H$ be a subgroup of $G$. Prove that $C_G(H) \le N_G(H)$.

\subsection*Problem S4.2 \textitLet $G$ be a cyclic group of order $n$. Prove that for each divisor $d$ of $n$, there exists exactly one subgroup of order $d$. \subsection*Exercise 4

If $|Z(G)| = p^2$, then $G$ is abelian. If $|Z(G)| = p$, then $G/Z(G)$ has order $p$, hence is cyclic. A well-known lemma states: if $G/Z(G)$ is cyclic, then $G$ is abelian. So $G$ is abelian in both cases. \endsolution

\newpage \section*Supplementary Problems for Chapter 4

\beginsolution Let $G = \langle g \rangle$, $|G|=n$. For $d \mid n$, write $n = dk$. Then $\langle g^k \rangle$ has order $d$. Uniqueness: if $H \le G$, $|H|=d$, then $H = \langle g^m \rangle$ where $g^m$ has order $d$, so $n / \gcd(n,m) = d$, implying $\gcd(n,m) = k$. But $\langle g^m \rangle = \langle g^\gcd(n,m) \rangle = \langle g^k \rangle$. So unique. \endsolution Then \[ ab = g^m g^n = g^m+n = g^n+m = g^n g^m = ba

\subsection*Exercise 4.3.12 \textitProve that if $H$ is the unique subgroup of a finite group $G$ of order $n$, then $H$ is normal in $G$.

% Custom commands \newcommand\Z\mathbbZ \newcommand\Q\mathbbQ \newcommand\R\mathbbR \newcommand\C\mathbbC \newcommand\F\mathbbF \newcommand\Aut\operatornameAut \newcommand\Inn\operatornameInn \newcommand\sgn\operatornamesgn \newcommand\ord\operatornameord \newcommand\lcm\operatornamelcm \renewcommand\phi\varphi

\subsection*Exercise 4.1.1 \textitProve that every cyclic group is abelian.

Divisors of 12: $1,2,3,4,6,12$. The subgroups are: \beginalign* &\langle 0 \rangle = \0\ \quad \text(order 1)\\ &\langle 6 \rangle = \0,6\ \quad \text(order 2)\\ &\langle 4 \rangle = \0,4,8\ \quad \text(order 3)\\ &\langle 3 \rangle = \0,3,6,9\ \quad \text(order 4)\\ &\langle 2 \rangle = \0,2,4,6,8,10\ \quad \text(order 6)\\ &\langle 1 \rangle = \Z_12 \quad \text(order 12) \endalign*