A question about the consequence of Sylow's theorems The Next CEO of Stack OverflowWikipedia article on Sylow's theoremsIs my textbook wrong about this corollary of Sylow's theorem?what does “the conjugacy part of sylow's theorems” denote?Question about soluble and cyclic groups of order pqtrouble applying Sylow's theoremsQuestion about Sylow subgroupsCounting the number of distinct elements in Sylow subgroups if $|G|=30$Simple Consequence of Sylow TheoremUnderstanding the third Sylow theoremSecond part of Sylow's Theorem about conjugacy
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A question about the consequence of Sylow's theorems
The Next CEO of Stack OverflowWikipedia article on Sylow's theoremsIs my textbook wrong about this corollary of Sylow's theorem?what does “the conjugacy part of sylow's theorems” denote?Question about soluble and cyclic groups of order pqtrouble applying Sylow's theoremsQuestion about Sylow subgroupsCounting the number of distinct elements in Sylow subgroups if $|G|=30$Simple Consequence of Sylow TheoremUnderstanding the third Sylow theoremSecond part of Sylow's Theorem about conjugacy
$begingroup$
The three Sylow theorems are stated here. I don't understand this statement "A consequence of theorem 3 is if $n_p = 1$, then the Sylow p-subgroup is a normal subgroup".
I understand that if $n_p = 1$, there exists only one Sylow p-subgroup, Say H, and by Theorem 2, $exists gin G$ s.t. $gHg^-1 = H$ but why is it true $forall gin G$ so that $H$ becomes normal? Thanks!
abstract-algebra group-theory sylow-theory
asked Mar 18 at 21:46
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
The three Sylow theorems are stated here. I don't understand this statement "A consequence of theorem 3 is if $n_p = 1$, then the Sylow p-subgroup is a normal subgroup".
I understand that if $n_p = 1$, there exists only one Sylow p-subgroup, Say H, and by Theorem 2, $exists gin G$ s.t. $gHg^-1 = H$ but why is it true $forall gin G$ so that $H$ becomes normal? Thanks!
abstract-algebra group-theory sylow-theory
asked Mar 18 at 21:46
manifoldedmanifolded
49519
$endgroup$
2
$begingroup$
It is trivially true that any conjugate of a Sylow $p$-subgroup is a Sylow $p$-subgroup (because it has the same order). Sylow's theorem states the non-trivial direction - that any two Sylow $p$-subgroups are conjugate.
$endgroup$
– vxnture
Mar 18 at 21:50
add a comment |
$begingroup$
The three Sylow theorems are stated here. I don't understand this statement "A consequence of theorem 3 is if $n_p = 1$, then the Sylow p-subgroup is a normal subgroup".
I understand that if $n_p = 1$, there exists only one Sylow p-subgroup, Say H, and by Theorem 2, $exists gin G$ s.t. $gHg^-1 = H$ but why is it true $forall gin G$ so that $H$ becomes normal? Thanks!
abstract-algebra group-theory sylow-theory
asked Mar 18 at 21:46
manifoldedmanifolded
49519
$endgroup$
The three Sylow theorems are stated here. I don't understand this statement "A consequence of theorem 3 is if $n_p = 1$, then the Sylow p-subgroup is a normal subgroup".
I understand that if $n_p = 1$, there exists only one Sylow p-subgroup, Say H, and by Theorem 2, $exists gin G$ s.t. $gHg^-1 = H$ but why is it true $forall gin G$ so that $H$ becomes normal? Thanks!
abstract-algebra group-theory sylow-theory
abstract-algebra group-theory sylow-theory
asked Mar 18 at 21:46
manifoldedmanifolded
49519
asked Mar 18 at 21:46
manifoldedmanifolded
49519
asked Mar 18 at 21:46
manifoldedmanifolded
49519
asked Mar 18 at 21:46
manifoldedmanifolded
49519
asked Mar 18 at 21:46
manifoldedmanifolded
49519
49519
2
$begingroup$
It is trivially true that any conjugate of a Sylow $p$-subgroup is a Sylow $p$-subgroup (because it has the same order). Sylow's theorem states the non-trivial direction - that any two Sylow $p$-subgroups are conjugate.
$endgroup$
– vxnture
Mar 18 at 21:50
add a comment |
2
$begingroup$
It is trivially true that any conjugate of a Sylow $p$-subgroup is a Sylow $p$-subgroup (because it has the same order). Sylow's theorem states the non-trivial direction - that any two Sylow $p$-subgroups are conjugate.
$endgroup$
– vxnture
Mar 18 at 21:50
2
2
$begingroup$
It is trivially true that any conjugate of a Sylow $p$-subgroup is a Sylow $p$-subgroup (because it has the same order). Sylow's theorem states the non-trivial direction - that any two Sylow $p$-subgroups are conjugate.
$endgroup$
– vxnture
Mar 18 at 21:50
$begingroup$
It is trivially true that any conjugate of a Sylow $p$-subgroup is a Sylow $p$-subgroup (because it has the same order). Sylow's theorem states the non-trivial direction - that any two Sylow $p$-subgroups are conjugate.
$endgroup$
– vxnture
Mar 18 at 21:50
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Any conjugate $gHg^-1$ of any subgroup is itself a subgroup of the same order. So if $gHg^-1 neq H$, we have a second Sylow $p$-subgroup of $G$, contradicting our assumption that there is only one such subgroup. Thus, $forall g in G, gHg^-1 = H text and H triangleleft G.$
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
$begingroup$
For each $gin G$ we have that $gHg^-1$ is a group of the same size as $H$, hence it is a $p$-Sylow subgroup as well. Since there is only one $p$-Sylow subgroup we conclude that $gHg^-1=H$ for all $gin G$. This means $H$ is normal in $G$.
Note that with the same proof we conclude there is a more general fact: if $G$ has only one subgroup of a specific finite size $d$ then this subgroup must be normal in $G$. The converse statement (that if $Htrianglelefteq G$ then there are no other subgroups of the same size as $H$) is obviously false in general. However, for Sylow subgroups the converse is also true and it follows from the second Sylow theorem.
answered Mar 18 at 21:54
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
From the theorem, $n_p=1$ implies
$$lvert G : N_G(P)rvert = 1,$$
where $N_G$ denotes the normalizer. This tells you that $N_G(P)$ is all of $G$ (in particular, since we're working with finite groups, $ lvert G : N_G(P)rvert = fraclvert G rvertlvert N_G(P)rvert$).
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
Any conjugate $gHg^-1$ of any subgroup is itself a subgroup of the same order. So if $gHg^-1 neq H$, we have a second Sylow $p$-subgroup of $G$, contradicting our assumption that there is only one such subgroup. Thus, $forall g in G, gHg^-1 = H text and H triangleleft G.$
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
$begingroup$
Any conjugate $gHg^-1$ of any subgroup is itself a subgroup of the same order. So if $gHg^-1 neq H$, we have a second Sylow $p$-subgroup of $G$, contradicting our assumption that there is only one such subgroup. Thus, $forall g in G, gHg^-1 = H text and H triangleleft G.$
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
$begingroup$
Any conjugate $gHg^-1$ of any subgroup is itself a subgroup of the same order. So if $gHg^-1 neq H$, we have a second Sylow $p$-subgroup of $G$, contradicting our assumption that there is only one such subgroup. Thus, $forall g in G, gHg^-1 = H text and H triangleleft G.$
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
$endgroup$
Any conjugate $gHg^-1$ of any subgroup is itself a subgroup of the same order. So if $gHg^-1 neq H$, we have a second Sylow $p$-subgroup of $G$, contradicting our assumption that there is only one such subgroup. Thus, $forall g in G, gHg^-1 = H text and H triangleleft G.$
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 21:51
Robert ShoreRobert Shore
3,603324
3,603324
add a comment |
add a comment |
$begingroup$
For each $gin G$ we have that $gHg^-1$ is a group of the same size as $H$, hence it is a $p$-Sylow subgroup as well. Since there is only one $p$-Sylow subgroup we conclude that $gHg^-1=H$ for all $gin G$. This means $H$ is normal in $G$.
Note that with the same proof we conclude there is a more general fact: if $G$ has only one subgroup of a specific finite size $d$ then this subgroup must be normal in $G$. The converse statement (that if $Htrianglelefteq G$ then there are no other subgroups of the same size as $H$) is obviously false in general. However, for Sylow subgroups the converse is also true and it follows from the second Sylow theorem.
answered Mar 18 at 21:54
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
For each $gin G$ we have that $gHg^-1$ is a group of the same size as $H$, hence it is a $p$-Sylow subgroup as well. Since there is only one $p$-Sylow subgroup we conclude that $gHg^-1=H$ for all $gin G$. This means $H$ is normal in $G$.
Note that with the same proof we conclude there is a more general fact: if $G$ has only one subgroup of a specific finite size $d$ then this subgroup must be normal in $G$. The converse statement (that if $Htrianglelefteq G$ then there are no other subgroups of the same size as $H$) is obviously false in general. However, for Sylow subgroups the converse is also true and it follows from the second Sylow theorem.
answered Mar 18 at 21:54
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
For each $gin G$ we have that $gHg^-1$ is a group of the same size as $H$, hence it is a $p$-Sylow subgroup as well. Since there is only one $p$-Sylow subgroup we conclude that $gHg^-1=H$ for all $gin G$. This means $H$ is normal in $G$.
Note that with the same proof we conclude there is a more general fact: if $G$ has only one subgroup of a specific finite size $d$ then this subgroup must be normal in $G$. The converse statement (that if $Htrianglelefteq G$ then there are no other subgroups of the same size as $H$) is obviously false in general. However, for Sylow subgroups the converse is also true and it follows from the second Sylow theorem.
answered Mar 18 at 21:54
MarkMark
10.4k1622
$endgroup$
For each $gin G$ we have that $gHg^-1$ is a group of the same size as $H$, hence it is a $p$-Sylow subgroup as well. Since there is only one $p$-Sylow subgroup we conclude that $gHg^-1=H$ for all $gin G$. This means $H$ is normal in $G$.
Note that with the same proof we conclude there is a more general fact: if $G$ has only one subgroup of a specific finite size $d$ then this subgroup must be normal in $G$. The converse statement (that if $Htrianglelefteq G$ then there are no other subgroups of the same size as $H$) is obviously false in general. However, for Sylow subgroups the converse is also true and it follows from the second Sylow theorem.
answered Mar 18 at 21:54
MarkMark
10.4k1622
answered Mar 18 at 21:54
MarkMark
10.4k1622
answered Mar 18 at 21:54
MarkMark
10.4k1622
answered Mar 18 at 21:54
MarkMark
10.4k1622
10.4k1622
add a comment |
add a comment |
$begingroup$
From the theorem, $n_p=1$ implies
$$lvert G : N_G(P)rvert = 1,$$
where $N_G$ denotes the normalizer. This tells you that $N_G(P)$ is all of $G$ (in particular, since we're working with finite groups, $ lvert G : N_G(P)rvert = fraclvert G rvertlvert N_G(P)rvert$).
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
$endgroup$
add a comment |
$begingroup$
From the theorem, $n_p=1$ implies
$$lvert G : N_G(P)rvert = 1,$$
where $N_G$ denotes the normalizer. This tells you that $N_G(P)$ is all of $G$ (in particular, since we're working with finite groups, $ lvert G : N_G(P)rvert = fraclvert G rvertlvert N_G(P)rvert$).
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
$endgroup$
add a comment |
$begingroup$
From the theorem, $n_p=1$ implies
$$lvert G : N_G(P)rvert = 1,$$
where $N_G$ denotes the normalizer. This tells you that $N_G(P)$ is all of $G$ (in particular, since we're working with finite groups, $ lvert G : N_G(P)rvert = fraclvert G rvertlvert N_G(P)rvert$).
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
$endgroup$
From the theorem, $n_p=1$ implies
$$lvert G : N_G(P)rvert = 1,$$
where $N_G$ denotes the normalizer. This tells you that $N_G(P)$ is all of $G$ (in particular, since we're working with finite groups, $ lvert G : N_G(P)rvert = fraclvert G rvertlvert N_G(P)rvert$).
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
answered Mar 18 at 21:55
Gary MoonGary Moon
89627
89627
add a comment |
add a comment |
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$begingroup$
It is trivially true that any conjugate of a Sylow $p$-subgroup is a Sylow $p$-subgroup (because it has the same order). Sylow's theorem states the non-trivial direction - that any two Sylow $p$-subgroups are conjugate.
$endgroup$
– vxnture
Mar 18 at 21:50