Dummit And Foote Solutions Chapter 4 Overleaf -
\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
\tableofcontents \newpage
\beginsolution Consider the action of $G$ on itself by left multiplication. This gives a homomorphism $\varphi: G \to S_2n$. However, a more refined approach uses Cayley's theorem and parity. Dummit And Foote Solutions Chapter 4 Overleaf
\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 \beginexercise[Section 4
\sectionConclusion and Further Directions
\sectionThe Class Equation and Consequences This gives a homomorphism $\varphi: G \to S_2n$
\beginsolution Fix $a \in A$. By transitivity, $A = \Orb(a)$. The Orbit-Stabilizer Theorem states: [ |\Orb(a)| = \fracG. ] Thus $|A| = |G| / |\Stab_G(a)|$, so $|A| \cdot |\Stab_G(a)| = |G|$. Hence $|A|$ divides $|G|$. \endsolution
