Dummit - And Foote Solutions Chapter 4 Overleaf

\sectionConclusion and Further Directions

\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

Alternatively, consider the action of $G$ on the set of all subsets of size $n$? A standard proof uses the regular representation and the sign homomorphism. Let $G$ act on itself by left multiplication; this yields an embedding $\pi: G \hookrightarrow S_2n$. Since $n$ is odd, $2n$ is even. Compose with the sign map $\sgn: S_2n \to \pm1$. The kernel of $\sgn \circ \pi$ is a subgroup of index at most $2$. If the image is $\pm1$, the kernel has index $2$ and hence order $n$. If the image is trivial, then every element acts as an even permutation. But in $S_2n$, a transposition is odd; careful analysis (see D&F) shows this forces a contradiction for $n$ odd. Thus the kernel is the desired subgroup of order $n$. \endsolution For $a \in \Fix(A)$, $|\Orb(a)| = 1$

\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 \endsolution Alternatively, consider the action of $G$ on

\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

\sectionGroup Actions and Permutation Representations