The exercises above illustrate the power of group actions in classifying finite groups, proving structural theorems (e.g., all groups of order $p^2$ are abelian), and laying the groundwork for the Sylow theorems. Mastery of Chapter 4 is essential for advanced topics such as representation theory, solvable groups, and the classification of finite simple groups.
\beginexercise[Section 4.5, Exercise 10] Prove that if $|G| = 12$, then $G$ has either one or four Sylow $3$-subgroups. \endexercise
\beginsolution Decompose $A$ into disjoint orbits. For any $a \notin \Fix(A)$, its orbit size is $|\Orb(a)| = |G|/|\Stab(a)|$. Since $G$ is a $p$-group, $|\Orb(a)|$ is a power of $p$ greater than $1$, hence divisible by $p$. For $a \in \Fix(A)$, $|\Orb(a)| = 1$. Therefore: [ |A| = \sum_\textorbits |\Orb(a)| = |\Fix(A)| + \sum_\textnon-fixed orbits (\textmultiple of p). ] Reducing modulo $p$ yields $|A| \equiv |\Fix(A)| \pmodp$. \endsolution
\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 Dummit And Foote Solutions Chapter 4 Overleaf
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\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
\beginthebibliography9 \bibitemDF Dummit, David S., and Richard M. Foote. \textitAbstract Algebra. 3rd ed., Wiley, 2004. \endthebibliography The exercises above illustrate the power of group
\sectionThe Orbit-Stabilizer Theorem
\sectionConclusion and Further Directions
\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 For $a \in \Fix(A)$, $|\Orb(a)| = 1$
\beginsolution Let $n_p$ and $n_q$ be the numbers of Sylow $p$- and $q$-subgroups. By Sylow, $n_p \equiv 1 \pmodp$ and $n_p \mid q$. Since $p \neq q$, $n_p = 1$ or $n_p = q$. Similarly, $n_q \equiv 1 \pmodq$ and $n_q \mid p^2$, so $n_q = 1, p, p^2$. If $n_p = 1$, the Sylow $p$-subgroup is normal and we are done. If $n_q = 1$, done. Assume $n_p = q$ and $n_q \neq 1$. Then $n_q = p$ or $p^2$. But $n_q \equiv 1 \pmodq$ forces $p \equiv 1 \pmodq$ or $p^2 \equiv 1 \pmodq$. These conditions contradict $p,q$ distinct and the counting of elements (each Sylow $q$-subgroup contributes $q-1$ non-identity elements, etc.). A standard counting argument shows $n_p = 1$ must hold. \endsolution
\beginsolution Fix $a \in A$. By transitivity, $A = \Orb(a)$. The Orbit-Stabilizer Theorem states: [ |\Orb(a)| = \frac. ] Thus $|A| = |G| / |\Stab_G(a)|$, so $|A| \cdot |\Stab_G(a)| = |G|$. Hence $|A|$ divides $|G|$. \endsolution
\beginexercise[Section 4.2, Exercise 2] Let $G$ act on a finite set $A$. Prove that if $G$ acts transitively on $A$, then $|A|$ divides $|G|$. \endexercise
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