Dummit And Foote Solutions Chapter 4 Overleaf Now

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

\sectionApplications to $p$-groups and Sylow Theorems Dummit And Foote Solutions Chapter 4 Overleaf

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

\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 Since $n$ is odd, $2n$ is even

\beginthebibliography9 \bibitemDF Dummit, David S., and Richard M. Foote. \textitAbstract Algebra. 3rd ed., Wiley, 2004. \endthebibliography