Let $p$ be a prime dividing $|G|$, and let $S_p$ be a $p$-Sylow subgroup of $G$. Show that...
up vote
0
down vote
favorite
Attempt: Trivially $N(S_p) subseteq N(N(S_p))$ because a subgroup is a subset of its normalizer. Then, let $gin N(N(S_p))$. Then we have $gN(S_p)g^{-1}=N(S_p)$ by the definition of normalizer. But $S_p unlhd N(S_p)$, and since $S_p$ is also a $p$-Sylow subgroup of $N(S_p), gS_pg^{-1}=S_p$, and so $gin N(S_p)$, then we are done. My question is whether it is okay I go from $S_p unlhd N(S_p)$ and $S_p$ a $p$-Sylow subgroup of $N(S_p)$ to it is conjugate to itself. This reminds me of if there is only one $p$-Sylow subgroup, it is equivalent of saying this $p$-Sylow subgroup is normal. Is the inverse statement of this true?
group-theory normal-subgroups sylow-theory
asked Nov 14 at 23:13
Decbelief
243
add a comment |
up vote
0
down vote
favorite
Attempt: Trivially $N(S_p) subseteq N(N(S_p))$ because a subgroup is a subset of its normalizer. Then, let $gin N(N(S_p))$. Then we have $gN(S_p)g^{-1}=N(S_p)$ by the definition of normalizer. But $S_p unlhd N(S_p)$, and since $S_p$ is also a $p$-Sylow subgroup of $N(S_p), gS_pg^{-1}=S_p$, and so $gin N(S_p)$, then we are done. My question is whether it is okay I go from $S_p unlhd N(S_p)$ and $S_p$ a $p$-Sylow subgroup of $N(S_p)$ to it is conjugate to itself. This reminds me of if there is only one $p$-Sylow subgroup, it is equivalent of saying this $p$-Sylow subgroup is normal. Is the inverse statement of this true?
group-theory normal-subgroups sylow-theory
asked Nov 14 at 23:13
Decbelief
243
Well, it is okay to go as you say but remarking very clearly that $;S_p;$ is a normal Sylow $;p,-$ subgroup of $;N_G(S_p);$ . And yes: a Sylow subgroup of a finite group is unique iff it is normal.
– DonAntonio
Nov 14 at 23:16
The statement is true even if $p$ does not divide $|G|$.
– C Monsour
Nov 15 at 0:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Attempt: Trivially $N(S_p) subseteq N(N(S_p))$ because a subgroup is a subset of its normalizer. Then, let $gin N(N(S_p))$. Then we have $gN(S_p)g^{-1}=N(S_p)$ by the definition of normalizer. But $S_p unlhd N(S_p)$, and since $S_p$ is also a $p$-Sylow subgroup of $N(S_p), gS_pg^{-1}=S_p$, and so $gin N(S_p)$, then we are done. My question is whether it is okay I go from $S_p unlhd N(S_p)$ and $S_p$ a $p$-Sylow subgroup of $N(S_p)$ to it is conjugate to itself. This reminds me of if there is only one $p$-Sylow subgroup, it is equivalent of saying this $p$-Sylow subgroup is normal. Is the inverse statement of this true?
group-theory normal-subgroups sylow-theory
asked Nov 14 at 23:13
Decbelief
243
Attempt: Trivially $N(S_p) subseteq N(N(S_p))$ because a subgroup is a subset of its normalizer. Then, let $gin N(N(S_p))$. Then we have $gN(S_p)g^{-1}=N(S_p)$ by the definition of normalizer. But $S_p unlhd N(S_p)$, and since $S_p$ is also a $p$-Sylow subgroup of $N(S_p), gS_pg^{-1}=S_p$, and so $gin N(S_p)$, then we are done. My question is whether it is okay I go from $S_p unlhd N(S_p)$ and $S_p$ a $p$-Sylow subgroup of $N(S_p)$ to it is conjugate to itself. This reminds me of if there is only one $p$-Sylow subgroup, it is equivalent of saying this $p$-Sylow subgroup is normal. Is the inverse statement of this true?
group-theory normal-subgroups sylow-theory
group-theory normal-subgroups sylow-theory
asked Nov 14 at 23:13
Decbelief
243
asked Nov 14 at 23:13
Decbelief
243
asked Nov 14 at 23:13
Decbelief
243
asked Nov 14 at 23:13
Decbelief
243
asked Nov 14 at 23:13
Decbelief
243
243
Well, it is okay to go as you say but remarking very clearly that $;S_p;$ is a normal Sylow $;p,-$ subgroup of $;N_G(S_p);$ . And yes: a Sylow subgroup of a finite group is unique iff it is normal.
– DonAntonio
Nov 14 at 23:16
The statement is true even if $p$ does not divide $|G|$.
– C Monsour
Nov 15 at 0:56
add a comment |
Well, it is okay to go as you say but remarking very clearly that $;S_p;$ is a normal Sylow $;p,-$ subgroup of $;N_G(S_p);$ . And yes: a Sylow subgroup of a finite group is unique iff it is normal.
– DonAntonio
Nov 14 at 23:16
The statement is true even if $p$ does not divide $|G|$.
– C Monsour
Nov 15 at 0:56
Well, it is okay to go as you say but remarking very clearly that $;S_p;$ is a normal Sylow $;p,-$ subgroup of $;N_G(S_p);$ . And yes: a Sylow subgroup of a finite group is unique iff it is normal.
– DonAntonio
Nov 14 at 23:16
Well, it is okay to go as you say but remarking very clearly that $;S_p;$ is a normal Sylow $;p,-$ subgroup of $;N_G(S_p);$ . And yes: a Sylow subgroup of a finite group is unique iff it is normal.
– DonAntonio
Nov 14 at 23:16
The statement is true even if $p$ does not divide $|G|$.
– C Monsour
Nov 15 at 0:56
The statement is true even if $p$ does not divide $|G|$.
– C Monsour
Nov 15 at 0:56
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998953%2flet-p-be-a-prime-dividing-g-and-let-s-p-be-a-p-sylow-subgroup-of-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Well, it is okay to go as you say but remarking very clearly that $;S_p;$ is a normal Sylow $;p,-$ subgroup of $;N_G(S_p);$ . And yes: a Sylow subgroup of a finite group is unique iff it is normal.
– DonAntonio
Nov 14 at 23:16
The statement is true even if $p$ does not divide $|G|$.
– C Monsour
Nov 15 at 0:56