HW5

(1) Suppose $1\ne N◃G$ and $G$ is nilpotent, then $N\cap C(G)\ne 1$.

\begin{proof} Since $G$ is nilpotent, $G$ is a direct product of Sylow subgroups. Assume that $G=P_1\times \cdots \times P_n$ where $P_i$ is the Sylow $p_i$-subgroups and $p_1,\cdots ,p_n$ are distinct primes. It follows that $N=Q_1\times \cdots \times Q_n$ where $Q_i⊲P_i$ are Sylow $p_i$-subgroups of $N$.

Note that $C(G)=C(P_1)\times \cdots \times C(P_n)$. If $N\cap C(G)=1$, then $Q_i\cap C(P_i)=1$ for all $i\in \{1,\cdots ,n\}$. By exercise (2) of HW3, for any $Q_i⊲P_i$ with $Q_i\cap C(P_i)=\{1\}$, we have that $Q_i=\{1\}$. It yields that $N=\{1\}$, contradiction. Therefore, $N\cap C(G)\ne 1$. \end{proof}

(2) Let $G$ be a nontrivial finite group.

• (a) If $G$ is solvable, then it contains a nontrivial normal abelian subgroup.
• (b) If $G$ is not solvable, then it contains a nontrivial normal subgroup $H$ such that $H^{\prime }=H$.

\begin{proof} First, we claim that $G^{(m)}$ is a normal subgroup of $G$ for any $m\in {\mathbb{N}}_{+}$. When $m=1$, $G^{(1)}=G^{\prime }=[G,G]$ and each element in $G^{\prime }$ can be written as $ab{a}^{-1}{b}^{-1}:=[a,b]$ with $a,b\in G$. For any $g\in G$, $[a,b{]}^g=[a^g,b^g]\in G$ and so $G^{(1)}⊲G$. Assume that $G^{(k)}⊲G$, then

$$G^{(k+1)}=[G^{(k)},G^{(k)}]=\{[a,b]:a,b\in G^{(k)}\}.$$

For any $g\in G$ and any $a,b\in G^{(k)}$, $[a,b{]}^g=[a^g,b^g]$. As $G^{(k)}⊲G$, we have that $a^g,b^g\in G^{(k)}$ and so $[a,b{]}^g\in G^{(k+1)}$ for any $g\in G$. Therefore, $G^{(k+1)}⊲G$. By induction hypothesis, we can show that $G^{(m)}⊲G$ for all $m\in {\mathbb{N}}_{+}$.

i) Since $G$ is solvable, there exists an $n$ such that $\{1\}\ne G^{(n)}⊲G$ and $G^{(n+1)}=\{1\}$. It follows that $(G^{(n)}{)}^{\prime }=\{1\}$ and so $G^{(n)}$ is abelian.

ii) Since $G$ is non-solvable, the length of derived series of $G$ is infinite. Therefore, there exists $n_2$ such that $(G^{(n_2)}{)}^{\prime }=G^{(n_2)}$ and this is the desired $H$. \end{proof}

(3) Any group of order $p^{2}q$ ( $p,q$ are distinct primes) is solvable.

\begin{proof} Assume that $p>q$. By the third Sylow theorem, $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid q$. It yields that $n_p=1$ and so $G$ has the unique Sylow $p$-subgroup $P$. Since $G/P$ is solvable by $|G/P|=q$ and $P$ is solvable, $G$ is also solvable.

Assume that $p<q$. By the third Sylow theorem, $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid q$. It follows that $n_p\in \{1,q\}$. Similarly, we get $n_q\in \{1,p^2\}$. Assume that $n_p=q$ and $n_q=p^2$. Since Sylow $q$-subgroup is cyclic, $G$ has $p^2$ Sylow $q$-subgroups yields that $G$ has $p^2(q-1)$ elements of order $q$. Hence, $G$ has $p^2-1$ elements of order $p^i(i=1,2)$, which contradicts with $n_p=q$. Therefore, at least one of $n_p,n_q$ equals $1$. If $n_p=1$ (rep. $n_q=1$), then $G$ has a solvable normal subgroup $N$ by $N$ being a $p$-group (rep. $q$-group). Further, note that $G/N$ is also solvable, it yields that $G$ is solvable. \end{proof}

(4) Is $A_4$ nilpotent? Show work.

\begin{proof} $A_4$ is not nilpotent.

Let $\omega =(12)(34)\in A_4$. Since $|G_{\omega }||{\omega }^G|=12$ and $|{\omega }^G|=3$, $G_{\omega }$ is of order $4$ and $G_{\omega }=\{1,(12)(34),(13)(24),(14)(23)\}$. As $Z(G)⩽G_{\omega }$ and $(123{)}^{\sigma }\ne (123)$ for any non-identity $\sigma \in G_{\omega }$, the center of $A_4$ is trivial. As each nilpotent group is a direct product of Sylow subgroups, its center is non-trivial. Therefore, $A_4$ is not nilpotent. \end{proof}

(5) Is $S_4$ nilpotent? Show work.

\begin{proof} $S_4$ is not nilpotent.

We claim that $Z(S_n)$ is trivial for any $n>2$. Let $\omega =(12\cdots n)$. Then $|S_n|=|C_{S_n}(\sigma )||{\sigma }^{S_n}|=|C_{S_n}(\sigma )|(n-1)!$ yields that $C_{S_n}=⟨\sigma ⟩$. Since $(12{)}^{\sigma }=(23)\ne (12)$, there is $\sigma \notin Z(G)$ and so $Z(S_n)=\{(1)\}$.

Therefore, $Z(S_4)=\{1\}$ and $S_4$ is not nilpotent. \end{proof}

(6) For what $n$ is $S_n$ nilpotent? Show work.

\begin{proof} By Exercise (5), for $n\in \{1,2\}$, $S_n$ is nilpotent; for $n>2$, $S_n$ is not nilpotent. \end{proof}