HW4

Let $G$ be a finite group of order $pm$ where $p>m$. Suppose $H<G$ is a subgroup and $|H|=p$. Then $H◃G$.

\begin{proof} Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$. It yields that $n_p=1$ by $p>m$ and so $G$ has the unique Sylow $p$-subgroup. Since $|H|=p$, then $H$ is the unique Sylow $p$-subgroup of $G$. By the second Sylow theorem, it follows that $H⊲G$. \end{proof}

Find all the Sylow 2-subgroups and Sylow 3-subgroups of $S_3$.

\begin{proof} Let $n_p$ be the number of Sylow $p$-subgroups of $S_3$. Since $|S_3|=6$, then $n_2=1\text{mbox}or3$ and $n_3=1$ by the third Sylow theorem. Note that both Sylow $2$-group and Sylow $3$-group are cyclic. As $⟨(12)⟩$ and $⟨(23)⟩$ are Sylow $2$-groups, we have that $n_2\ne 1$ and so $n_2=3$. Therefore, all the Sylow $2$-subgroups are

$$⟨(12)⟩,⟨(23)⟩,⟨(13)⟩,$$

and the unique Sylow $3$-subgroup is $⟨(123)⟩$. \end{proof}

Find all the Sylow 2-subgroups and Sylow 3-subgroups of $S_4$.

\begin{proof} Let $n_p$ be the number of Sylow $p$-subgroups of $S_4$. Since $|S_4|=24$, then $n_2=1\text{mbox}or3$ and $n_3=1\text{mbox}or4$ by the third Sylow theorem.

Note that $⟨(1234),(13)⟩$ is a Sylow $2$-subgroup and it is not normal. It yields that $n_3\ne 1$ and so $n_2=3$. Therefore, all Sylow $2$-subgroups are

$$⟨(1234),(13)⟩,⟨(1324),(12)⟩,⟨(1243),(14)⟩.$$

Since Sylow $3$-subgroups are cyclic, all Sylow $3$-subgroups are $$ \left\langle(123)\right\rangle,\left\langle(124)\right\rangle,\left\langle(134)\right\rangle,\left\langle(234)\right\rangle

and $n_3=4$. `\end{proof}` **Find all the Sylow 2-subgroups and Sylow 3-subgroups of $S_5$.** `\begin{proof}` Let $n_p$ be the number of Sylow $p$-subgroups of $S_5$. Since $|S_5|=120$, then $n_2\in\{1,3,5,15\}$ and $n_3\in\{1,4,10,40\}$ by the third Sylow theorem. Note that $\left\langle(1234),(13)\right\rangle<S_5$ is a Sylow $2$-subgroup of $S_5$. Let $\mathcal S=\{\left\langle(1234),(13)\right\rangle ,\left\langle(1324),(12)\right\rangle ,\left\langle(1243),(14)\right\rangle\}$, and let $\mathcal S^g=\{\left\langle(1234)^g,(13)^g\right\rangle ,\left\langle(1324)^g,(12)^g\right\rangle ,\left\langle(1243)^g,(14)^g\right\rangle\}$ for any $g\in S_5$. Then $\mathcal S,\mathcal{S}^{(15)},\mathcal{S}^{(25)},\mathcal{S}^{(35)},\mathcal{S}^{(45)}$ are sets of Sylow $2$-subgroups which intersects trivially. Therefore, $n_2=15$ and the set of all Sylow $2$-subgroups is

\mathcal S\sqcup\mathcal{S}^{(15)}\sqcup\mathcal{S}^{(25)}\sqcup\mathcal{S}^{(35)}\sqcup\mathcal{S}^{(45)}\mbox{, where }\mathcal S={\left\langle(1234),(13)\right\rangle ,\left\langle(1324),(12)\right\rangle ,\left\langle(1243),(14)\right\rangle}.

Since the order of a Sylow $3$-subgroup is $3$, then each of Sylow $3$-subgroup is generated by an element of order $3$. Note that there are ${5\choose 3}=10$ elements of order $3$ in $S_5$. Therefore, $n_3=10$ and the set of all Sylow $3$-subgroups is

\big{\left\langle(ijk)\right\rangle :{i,j,k}\subseteq{1,2,3,4,5},i<j<k\big}.

`\end{proof}` **Suppose $|G|=p^n q$ where $p>q$ are both primes. Then $G$ contains a unique normal subgroup of index $q$.** `\begin{proof}` By the third Sylow theorem, the number of Sylow $p$-subgroups satisfies that $n_p\equiv1\pmod{p}$ and $n_p\mid q$. Since $q$ is a prime, then $n_p\in\{1,q\}$. Assume that $n_q=1$, then $q\equiv 1\pmod p$. Then it yields that $q-1=kp>kq$ for some $k\in \mathbb{N}$ and so $k=0$, $q=1$, which is impossible. Therefore, $n_p=1$ and so $G$ has the unique Sylow $p$-subgroup $P$. By the second Sylow theorem, $P\lhd G$ is the unique normal subgroup of $G$ of index $q$. `\end{proof}` **If $|G|=56$, then it contains a normal Sylow subgroup.** `\begin{proof}` Let $n_7$ be the number of Sylow $7$-subgroups of $G$. By the third Sylow theorem, $n_7\equiv 1\pmod 7$ and $n_7\mid8$. It follows that $n_7=1\mbox{ or }8$. Since each Sylow $7$-subgroup is cyclic, then for any two distinct Sylow $7$-subgroups $P_1$ and $P_2$ we have that $P_1\cap P_2=\{e_G\}$. If $n_7=8$, then $G$ has $n_7\times 6=48$ elements of order $7$ and so $G$ has at most $8$ elements of order $2^k,k\in \mathbb{N}$. Therefore, $G$ has the unique Sylow $2$-subgroup $P$ and $P\lhd G$. If $n_7=1$, then $G$ has the unique Sylow $7$-subgroup $Q$ and $Q\lhd G$. `\end{proof}` **If $|G|=200$, then it contains a normal Sylow subgroup.** `\begin{proof}` Let $n_5$ be the number of Sylow $5$-subgroups of $G$. By the third Sylow theorem, $n_5\equiv 1\pmod 5$ and $n_5\mid8$. It follows that $n_5=1$ and so $G$ has the unique normal Sylow $5$-subgroup. `\end{proof}`