\tableofcontents \newpage
\beginsolution Let $n_3$ denote the number of Sylow $3$-subgroups. By Sylow's theorems, $n_3 \equiv 1 \pmod3$ and $n_3 \mid 4$. The divisors of $4$ are $1,2,4$. Which are $\equiv 1 \pmod3$? $1 \equiv 1 \pmod3$, $4 \equiv 1 \pmod3$, but $2 \equiv 2 \pmod3$. Hence $n_3 = 1$ or $n_3 = 4$. No other possibilities. \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
\beginexercise[Section 4.4, Exercise 12] Let $G$ be a group of order $p^2q$ with $p$ and $q$ distinct primes. Prove that $G$ has a normal Sylow subgroup. \endexercise
\beginexercise[Section 4.5, Exercise 10] Prove that if $|G| = 12$, then $G$ has either one or four Sylow $3$-subgroups. \endexercise Which are $\equiv 1 \pmod3$
\sectionThe Orbit-Stabilizer Theorem
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\sectionGroup Actions on Sylow Subgroups
\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 No other possibilities