If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself
Problem 117
Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.
Then show that $N_G(H)=H$.
Add to solve later
Sponsored Links
Hint.
Use the conjugate part of the Sylow theorem.
See the second statement of the Sylow theorem.
Proof.
It is clear that $H \subset N_G(H)$.
So we show that $N_G(H) \subset H$.
Take any $a \in N_G(H)$. Since $P \subset N_G(P) \subset H$, we have
\[aPa^{-1} \subset aHa^{-1}=H,\]
where the last step follows from $a \in N_G(H)$.
It follows that $P$ and $aPa^{-1}$ are both Sylow subgroups in $H$.
By the Sylow theorem, any two $p$-Sylow subgroups are conjugate. Thus there exists $b \in H$ such that $bPb^{-1}=aPa^{-1}$.
This implies that $(b^{-1}a)P(b^{-1}a)^{-1}=P$ and thus $b^{-1}a \in N_G(P) \subset H$. Hence we have $a \in H$ since $b \in H$.
This shows that $N_G(H) \subset H$, hence $N_G(H)=H$ as required.
Corollary (The Normalizer of the Normalizer of a Sylow subgroup)
We apply the result to the case $H=N_G(P)$, and obtain the following result.
That is, we have
\[N_G(N_G(P))=N_G(P).\]
Add to solve later
Sponsored Links
More from my site
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 $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]- Sylow Subgroups of a Group of Order 33 is Normal Subgroups
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]
- Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]
- Subgroup Containing All $p$-Sylow Subgroups of a Group
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 […]
Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57
Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.
Then determine the number of elements in $G$ of order $3$.
Proof.
Observe the prime factorization $57=3\cdot 19$.
Let $n_{19}$ be the number of Sylow $19$-subgroups of $G$.
By […]- Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
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 […]
Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4
Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.
Hint.
Use Sylow's theorem.
(See Sylow’s Theorem (Summary) for a review of Sylow's theorem.)
Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]- If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse
Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element.
Show that $x$ is not conjugate to $x^{-1}$.
Proof.
Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$.
Then we have
\[xg=gx^{-1}. […]
