Dummit And Foote Solutions Chapter 4 Overleaf High Quality -

\beginsolution Let $|G| = p^2$. The center $Z(G)$ is nontrivial by the class equation (since $|G| = |Z(G)| + \sum |G:C_G(g_i)|$, each term divisible by $p$). So $|Z(G)| = p$ or $p^2$.

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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

\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

\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 $d$. \beginsolution Let $|G| = p^2$

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

\beginsolution Let $|H| = n$ and suppose $H$ is the only subgroup of $G$ with order $n$. For any $g \in G$, consider $gHg^-1$. Conjugation is an automorphism of $G$, so $|gHg^-1| = |H| = n$. Thus $gHg^-1$ is also a subgroup of $G$ of order $n$. By uniqueness, $gHg^-1 = H$ for all $g \in G$. Hence $H \trianglelefteq G$. \endsolution Then $\langle g^k \rangle$ has order $d$

% 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