Dummit And Foote Solutions Chapter 4 Overleaf High Quality Online

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\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

\newpage \section*Supplementary Problems for Chapter 4 Dummit And Foote Solutions Chapter 4 Overleaf High Quality

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

\title\textbfDummit \& Foote \textitAbstract Algebra \\ Chapter 4 Solutions \authorYour Name \date\today Thus $gHg^-1$ is also a subgroup of $G$ of order $n$

\subsection*Exercise 4.5.9 \textitG:H

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 so $|gHg^-1| = |H| = n$.

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