\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 4.2, Exercise 2] Let $G$ act on a finite set $A$. Prove that if $G$ acts transitively on $A$, then $|A|$ divides $|G|$. \endexercise
\sectionGroup Actions on Sylow Subgroups Dummit And Foote Solutions Chapter 4 Overleaf
\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
\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
\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 4.2, Exercise 8] Let $G$ be a $p$-group acting on a finite set $A$. Prove that [ |A| \equiv |\Fix(A)| \pmodp, ] where $\Fix(A) = a \in A : g \cdot a = a \text for all g \in G$. \endexercise For $a \in \Fix(A)$, $|\Orb(a)| = 1$
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\beginsolution Let $G$ act on $G/H = gH : g \in G$ by $g \cdot (xH) = (gx)H$. \beginenumerate \item \textbfTransitivity: Take any two cosets $aH, bH \in G/H$. Choose $g = ba^-1 \in G$. Then [ g \cdot (aH) = (ba^-1a)H = bH. ] Hence, there is exactly one orbit, so the action is transitive. \item \textbfStabilizer of $1H$: [ \Stab_G(1H) = g \in G : g \cdot (1H) = 1H = g \in G : gH = H. ] But $gH = H$ if and only if $g \in H$. Therefore $\Stab_G(1H) = H$. \endenumerate \endsolution
\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