Dummit And Foote Solutions Chapter 4 Overleaf Page

\documentclass[12pt, leqno]article \usepackage[utf8]inputenc \usepackageamsmath, amssymb, amsthm, amscd \usepackage[margin=1in]geometry \usepackageenumitem \usepackagetitlesec \usepackagexcolor % -------------------------------------------------------------- % Custom Commands for Dummit & Foote Notation % -------------------------------------------------------------- \newcommand\Z\mathbbZ \newcommand\R\mathbbR \newcommand\C\mathbbC \newcommand\Q\mathbbQ \newcommand\F\mathbbF \newcommand\Stab\textStab \newcommand\Fix\textFix \newcommand\Orb\textOrb \newcommand\sgn\textsgn \newcommand\Aut\textAut \newcommand\Inn\textInn \newcommand\soc\textSoc \newcommand\Ker\textKer \newcommand\Image\textIm

\beginsolution Recall that $Z(G)$ is nontrivial for any $p$-group. Thus $|Z(G)| = p$ or $p^2$. If $|Z(G)| = p^2$, done. Suppose $|Z(G)| = p$. Then $G/Z(G)$ has order $p$, hence cyclic. A standard theorem states: if $G/Z(G)$ is cyclic, then $G$ is abelian. This contradicts $|Z(G)| = p < p^2$. Hence $|Z(G)| \neq p$, so $|Z(G)| = p^2$ and $G$ is abelian. \endsolution Dummit And Foote Solutions Chapter 4 Overleaf

You can copy and paste this code directly into a new Overleaf project. Suppose $|Z(G)| = p$

\beginexercise[Section 4.3, Exercise 15] Let $G$ be a $p$-group and let $N$ be a nontrivial normal subgroup of $G$. Prove that $N \cap Z(G) \neq 1$. \endexercise This contradicts $|Z(G)| = p < p^2$

\beginsolution Let $H = N_G(P)$. By definition, $P \triangleleft H$ (since $P$ is normal in its normalizer). Hence $P$ is the unique Sylow $p$-subgroup of $H$. Now let $g \in N_G(H)$. Then $gPg^-1 \subseteq gHg^-1 = H$, so $gPg^-1$ is also a Sylow $p$-subgroup of $H$. By uniqueness, $gPg^-1 = P$. Thus $g \in N_G(P) = H$. Therefore $N_G(H) \subseteq H$, and the reverse inclusion is trivial. So $N_G(H) = H$. \endsolution

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