Dummit And Foote Solutions Chapter 4 Overleaf ⟶ < SIMPLE >

\beginsolution Consider the action of $G$ on $N$ by conjugation. Since $N \triangleleft G$, this action is well-defined. The fixed points of this action are $N \cap Z(G)$. By the $p$-group fixed point theorem (Exercise 4.2.8), $|N| \equiv |N \cap Z(G)| \pmodp$. Since $|N|$ is a power of $p$ and $N$ is nontrivial, $p \mid |N|$. Hence $p \mid |N \cap Z(G)|$, so $|N \cap Z(G)| \geq p > 1$. Thus $N \cap Z(G) \neq 1$. \endsolution

% Theorem environments \newtheoremtheoremTheorem[section] \newtheoremlemma[theorem]Lemma \newtheoremproposition[theorem]Proposition \newtheoremcorollary[theorem]Corollary \theoremstyledefinition \newtheoremdefinition[theorem]Definition \newtheoremexample[theorem]Example \newtheoremexerciseExercise[section] \newtheoremsolutionSolution[section] Dummit And Foote Solutions Chapter 4 Overleaf

\sectionGroup Actions on Sylow Subgroups \beginsolution Consider the action of $G$ on $N$

\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 By the $p$-group fixed point theorem (Exercise 4

\beginexercise[Section 4.4, Exercise 6] Prove that if $|G| = p^n$ for $p$ prime and $n \geq 1$, then $Z(G)$ is nontrivial. \endexercise

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