Subgroup Containing All $p$-Sylow Subgroups of a Group
Problem 227
Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.
Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.
Then show that $N$ contains all $p$-Sylow subgroups of $G$.
Since $p$ does not divide the index $|G: N|$, the order of $N$ is of the form
\[|N|=p^am, \text{ where } m|n.\]
By Sylow’s theorem, the group $N$ has a $p$-Sylow subgroup $P$.
Since the order of $P$ is $p^a$ and $P$ is a subgroup of $G$, it is also a Sylow subgroup of $G$.
Let $P’$ be any $p$-Sylow subgroup of $G$. Then by Sylow’s theorem, two $p$-Sylow subgroups are conjugate.
Thus there exists $g\in G$ such that
\[g^{-1}Pg=P’.\]
Then since $N$ is normal in $G$, we have
\begin{align*}
P’=g^{-1}Pg< g^{-1}Ng=N
\end{align*}
and $P'$ is a subgroup of $N$.
Proof 2.
Let $P$ be any $p$-Sylow subgroup of $G$. Seeking a contradiction, assume that $P \not \subset N$. Thus there exists $x \in P \setminus N$.
Let $H=\langle x \rangle$ be a subgroup of $G$ generated by $x$.
Since the order of $x$ is a power of $p$, the order of $H$ is a power of $p$.
Since $N \triangleleft G$, by the second isomorphism theorem, we have
\[|HN :N|=| H: H \cap N| \tag{*}.\]
Also, the chain of subgroups $N < HN < G$ implies
\[|G: N|=|G: HN| |HN: N|.\]
Combining this with (*), we obtain
\begin{align*}
|G: N|=|G: HN|| H: H \cap N|.
\end{align*}
Since $H\not \subset N$, $| H: H \cap N|$ is a positive power of $p$.
However, this contradicts that $p$ does not divide $|G: N|$.
Hence $P$ must be contained in $N$.
If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
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Hint.
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If a group $G$ has order $pq$, then show the followings.
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(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]
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Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]
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Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]
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Proof.
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Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
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Show that any subgroup of index $2$ in a group is a normal subgroup.
Hint.
Left (right) cosets partition the group into disjoint sets.
Consider both left and right cosets.
Proof.
Let $H$ be a subgroup of index $2$ in a group $G$.
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Let $G$ be a finite group and let $N$ be a normal subgroup of $G$.
Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.
(a) Prove that $N=\{a\in G \mid a^n=e\}$.
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Proof.
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