\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
\beginexercise[Section 4.5, Exercise 10] Prove that if $|G| = 12$, then $G$ has either one or four Sylow $3$-subgroups. \endexercise
\sectionThe Orbit-Stabilizer Theorem
\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
\beginabstract This document presents rigorous solutions to selected exercises from Chapter 4 of Dummit and Foote's \textitAbstract Algebra, Third Edition. The focus is on group actions, orbit-stabilizer theorem, $p$-groups, and applications to Sylow theory. Each solution emphasizes clear reasoning and formal justification. \endabstract Dummit And Foote Solutions Chapter 4 Overleaf
\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
\beginexercise[Section 4.3, Exercise 11] Let $G$ be a group of order $p^2$ where $p$ is prime. Prove that $G$ is abelian. \endexercise \beginsolution Recall that $Z(G)$ is nontrivial for any
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