HW3

In the following, G is always a finite group.

(1) $G$ is a finite group of order $p^n,N◃G$ such that $|N|=p$. Then $N$ is contained in the center of $G$.

\begin{proof} Assume that there exists $n_0\in N$ such that $n\notin C(G)$. Define $n_0^G:=\{g^{-1}{n}_{0}g:\forall g\in G\}$. Then $|n_0^G||C_G(n_0)|=|G|=p^n$. Since $n_0\notin C(G)$, then $C_G(n)\ne G$, $|n^G|\ne 1$ and $|n^G|⩾p$. As $N$ is a normal subgroup of $G$, then $|n^G|⩽|N|=p$ and so $n_0^G=N$. It is impossible because $e_G\notin n_0^G$. Therefore, $N$ is contained in the center of $G$.

Alternating proof. By NC lemma, $G/C_G(N)⩽\mathrm{Aut}(C_p)=C_{p-1}$ and so $G/C_G(N)=1$. Thus $G=C_G(N)$ and $N⩽Z(G)$. \end{proof}

(2) If $G$ is a finite $p$-group and $\left\{e_G\right\}\ne H◃G$, then $H\cap C(G)\ne \left\{e_G\right\}$. (I always use $\left\{e_G\right\}$ to denote the trivial subgroup).

\begin{proof} Assume that $H\cap C(G)=\{e_G\}$. It follows that for any non-identity element $h\in H$, $h^G:=\{g^{-1}hg:\forall g\in G\}$ is a subset of $H$ with $|h^G|\ne 1$. Consider the group action $G$ acting on $H$ by conjugation, then there exist $h_1,\cdots ,h_m\in G$ such that $H=\{e_G\}\cup h_1^G\cup \cdots \cup h_m^G$. Thus, $|H|=1+|h_1^G|+\cdots +|h_m^G|$. As $|h_i^G|=|G|/|C_G(h_i)|=p^{i_m}$ for some $i_m\in {\mathbb{N}}_{+}$, we have that $|H|\equiv 1\mathrm{mod}p$, which contradicts with $H⩽G$ and $G$ $p$-group. \end{proof}

(3) Suppose $P$ is a normal Sylow $p$-subgroup of $G$. If $f:G\to G$ is an endomorphism, show $f(P)<P$.

\begin{proof} For any $g\in P$, the order of $g$ is $p^k$ for some $k\in \mathbb{N}$. Since $f(g{)}^{p^k}=f(g^{p^k})=f(e_G)=e_G$, the order of $f(g)$ is $p^{\ell }$ for some $\ell ⩽k$. Therefore, $f(P)$ is a $p$-group. By second Sylow theorem, $f(P)$ is a subgroup for some Sylow $p$-subgroup. As $P$ is a normal Sylow $p$-subgroup of $G$, by second Sylow theorem $P$ is the unique Sylow $p$-subgroup of $G$ and so $f(P)⩽P$. \end{proof}

(4) Suppose $H◃G$ and $|H|=p^k$. Then $H$ is contained in every Sylow $p$-subgroup of $G$.

\begin{proof} Since $H$ is a $p$-group, for any Sylow $p$-subgroup $P$, there exists $g\in G$ such that $g^{-1}Hg⩽P$. Since $H⊲G$, then $g^{-1}Hg=H$ and so $H$ is contained in every Sylow $p$-subgroup of $G$. \end{proof}