\beginsolution Apply the class equation: [ |G| = |Z(G)| + \sum_i [G : C_G(g_i)], ] where the sum runs over non-central conjugacy classes. Each $[G : C_G(g_i)] > 1$ is a power of $p$ (since $C_G(g_i)$ is a subgroup). Thus $p$ divides each term in the sum. Also $p \mid |G|$. Hence $p \mid |Z(G)|$. Therefore $|Z(G)| \geq p$, so $Z(G)$ is nontrivial. \endsolution
\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 Dummit And Foote Solutions Chapter 4 Overleaf
% Custom colors for clarity \definecolornoteRGB0,100,0 \beginsolution Apply the class equation: [ |G| =
\sectionConclusion and Further Directions Also $p \mid |G|$
