Counterexample for the normalizer being a normal subgroup
$begingroup$
Let $G$ be a group, and $H$ a subgroup.
$H$'s normalizer is defined: $N(H):={gin G| gHg^{-1}=H }$.
Prove $N(H)$ is a normal subgroup of G, or give counterexample.
Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.
Thans in advance for any assistance!
abstract-algebra group-theory examples-counterexamples normal-subgroups
edited Dec 27 '18 at 17:08
Shaun
9,444113684
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group, and $H$ a subgroup.
$H$'s normalizer is defined: $N(H):={gin G| gHg^{-1}=H }$.
Prove $N(H)$ is a normal subgroup of G, or give counterexample.
Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.
Thans in advance for any assistance!
abstract-algebra group-theory examples-counterexamples normal-subgroups
edited Dec 27 '18 at 17:08
Shaun
9,444113684
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group, and $H$ a subgroup.
$H$'s normalizer is defined: $N(H):={gin G| gHg^{-1}=H }$.
Prove $N(H)$ is a normal subgroup of G, or give counterexample.
Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.
Thans in advance for any assistance!
abstract-algebra group-theory examples-counterexamples normal-subgroups
edited Dec 27 '18 at 17:08
Shaun
9,444113684
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
$endgroup$
Let $G$ be a group, and $H$ a subgroup.
$H$'s normalizer is defined: $N(H):={gin G| gHg^{-1}=H }$.
Prove $N(H)$ is a normal subgroup of G, or give counterexample.
Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.
Thans in advance for any assistance!
abstract-algebra group-theory examples-counterexamples normal-subgroups
abstract-algebra group-theory examples-counterexamples normal-subgroups
edited Dec 27 '18 at 17:08
Shaun
9,444113684
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
edited Dec 27 '18 at 17:08
Shaun
9,444113684
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
edited Dec 27 '18 at 17:08
Shaun
9,444113684
edited Dec 27 '18 at 17:08
Shaun
9,444113684
edited Dec 27 '18 at 17:08
Shaun
9,444113684
9,444113684
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
asked Apr 3 '14 at 14:47
dsfsfdsfsf
377311
377311
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
$endgroup$
add a comment |
$begingroup$
An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
$endgroup$
1
$begingroup$
Because these are (normalizers of) non-normal Sylow subgroups. ;)
$endgroup$
– Dune
Apr 3 '14 at 15:19
1
$begingroup$
@Dune That's a bit circular, though, eh? ;)
$endgroup$
– Alexander Gruber♦
Apr 3 '14 at 15:41
$begingroup$
Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned.
$endgroup$
– Dune
Apr 3 '14 at 20:16
add a comment |
$begingroup$
For any finite group $G$ and any $p$-Sylow subgroup $P leq G$ the following holds:
Every subgroup $U$ with $N_G(P) leq U leq G$ satisfies $N_G(U) = U$.
Especially every normalizer $N_G(P)$ provides a counterexample unless $P$ is normal in $G$. Can you think of examples of Sylow subgroups which are not normal?
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
$endgroup$
add a comment |
$begingroup$
Let $G=langle f,r : f^2 = r^8 = 1, rf =fr^7 rangle$ be the dihedral group of order 16, and let $H=langle f rangle$. Then $N_G(H) = langle f,r^4 rangle$ and $N_G( N_G(H) ) = langle f, r^2 rangle$ and $N_G( N_G( N_G(H) ) ) = langle f,r rangle =G$.
In other words, $H$ is not normal, neither is its normalizer, but the normalizer of the normalizer is normal.
Using dihedral groups of order $2^n$ you can create arbitrarily long chains of non-normal normalizers that eventually end up being normal.
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
$endgroup$
add a comment |
$begingroup$
The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
$endgroup$
add a comment |
$begingroup$
The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
$endgroup$
The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
answered Apr 3 '14 at 15:13
ah11950ah11950
2,100517
2,100517
add a comment |
add a comment |
$begingroup$
An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
$endgroup$
1
$begingroup$
Because these are (normalizers of) non-normal Sylow subgroups. ;)
$endgroup$
– Dune
Apr 3 '14 at 15:19
1
$begingroup$
@Dune That's a bit circular, though, eh? ;)
$endgroup$
– Alexander Gruber♦
Apr 3 '14 at 15:41
$begingroup$
Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned.
$endgroup$
– Dune
Apr 3 '14 at 20:16
add a comment |
$begingroup$
An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
$endgroup$
1
$begingroup$
Because these are (normalizers of) non-normal Sylow subgroups. ;)
$endgroup$
– Dune
Apr 3 '14 at 15:19
1
$begingroup$
@Dune That's a bit circular, though, eh? ;)
$endgroup$
– Alexander Gruber♦
Apr 3 '14 at 15:41
$begingroup$
Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned.
$endgroup$
– Dune
Apr 3 '14 at 20:16
add a comment |
$begingroup$
An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
$endgroup$
An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
edited Apr 3 '14 at 21:12
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
answered Apr 3 '14 at 15:16
Alexander Gruber♦Alexander Gruber
20k25103174
20k25103174
1
$begingroup$
Because these are (normalizers of) non-normal Sylow subgroups. ;)
$endgroup$
– Dune
Apr 3 '14 at 15:19
1
$begingroup$
@Dune That's a bit circular, though, eh? ;)
$endgroup$
– Alexander Gruber♦
Apr 3 '14 at 15:41
$begingroup$
Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned.
$endgroup$
– Dune
Apr 3 '14 at 20:16
add a comment |
1
$begingroup$
Because these are (normalizers of) non-normal Sylow subgroups. ;)
$endgroup$
– Dune
Apr 3 '14 at 15:19
1
$begingroup$
@Dune That's a bit circular, though, eh? ;)
$endgroup$
– Alexander Gruber♦
Apr 3 '14 at 15:41
$begingroup$
Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned.
$endgroup$
– Dune
Apr 3 '14 at 20:16
1
1
$begingroup$
Because these are (normalizers of) non-normal Sylow subgroups. ;)
$endgroup$
– Dune
Apr 3 '14 at 15:19
$begingroup$
Because these are (normalizers of) non-normal Sylow subgroups. ;)
$endgroup$
– Dune
Apr 3 '14 at 15:19
1
1
$begingroup$
@Dune That's a bit circular, though, eh? ;)
$endgroup$
– Alexander Gruber♦
Apr 3 '14 at 15:41
$begingroup$
@Dune That's a bit circular, though, eh? ;)
$endgroup$
– Alexander Gruber♦
Apr 3 '14 at 15:41
$begingroup$
Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned.
$endgroup$
– Dune
Apr 3 '14 at 20:16
$begingroup$
Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned.
$endgroup$
– Dune
Apr 3 '14 at 20:16
add a comment |
$begingroup$
For any finite group $G$ and any $p$-Sylow subgroup $P leq G$ the following holds:
Every subgroup $U$ with $N_G(P) leq U leq G$ satisfies $N_G(U) = U$.
Especially every normalizer $N_G(P)$ provides a counterexample unless $P$ is normal in $G$. Can you think of examples of Sylow subgroups which are not normal?
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
$endgroup$
add a comment |
$begingroup$
For any finite group $G$ and any $p$-Sylow subgroup $P leq G$ the following holds:
Every subgroup $U$ with $N_G(P) leq U leq G$ satisfies $N_G(U) = U$.
Especially every normalizer $N_G(P)$ provides a counterexample unless $P$ is normal in $G$. Can you think of examples of Sylow subgroups which are not normal?
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
$endgroup$
add a comment |
$begingroup$
For any finite group $G$ and any $p$-Sylow subgroup $P leq G$ the following holds:
Every subgroup $U$ with $N_G(P) leq U leq G$ satisfies $N_G(U) = U$.
Especially every normalizer $N_G(P)$ provides a counterexample unless $P$ is normal in $G$. Can you think of examples of Sylow subgroups which are not normal?
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
$endgroup$
For any finite group $G$ and any $p$-Sylow subgroup $P leq G$ the following holds:
Every subgroup $U$ with $N_G(P) leq U leq G$ satisfies $N_G(U) = U$.
Especially every normalizer $N_G(P)$ provides a counterexample unless $P$ is normal in $G$. Can you think of examples of Sylow subgroups which are not normal?
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
edited Apr 3 '14 at 15:11
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
answered Apr 3 '14 at 15:01
DuneDune
4,46711231
4,46711231
add a comment |
add a comment |
$begingroup$
Let $G=langle f,r : f^2 = r^8 = 1, rf =fr^7 rangle$ be the dihedral group of order 16, and let $H=langle f rangle$. Then $N_G(H) = langle f,r^4 rangle$ and $N_G( N_G(H) ) = langle f, r^2 rangle$ and $N_G( N_G( N_G(H) ) ) = langle f,r rangle =G$.
In other words, $H$ is not normal, neither is its normalizer, but the normalizer of the normalizer is normal.
Using dihedral groups of order $2^n$ you can create arbitrarily long chains of non-normal normalizers that eventually end up being normal.
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
$endgroup$
add a comment |
$begingroup$
Let $G=langle f,r : f^2 = r^8 = 1, rf =fr^7 rangle$ be the dihedral group of order 16, and let $H=langle f rangle$. Then $N_G(H) = langle f,r^4 rangle$ and $N_G( N_G(H) ) = langle f, r^2 rangle$ and $N_G( N_G( N_G(H) ) ) = langle f,r rangle =G$.
In other words, $H$ is not normal, neither is its normalizer, but the normalizer of the normalizer is normal.
Using dihedral groups of order $2^n$ you can create arbitrarily long chains of non-normal normalizers that eventually end up being normal.
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
$endgroup$
add a comment |
$begingroup$
Let $G=langle f,r : f^2 = r^8 = 1, rf =fr^7 rangle$ be the dihedral group of order 16, and let $H=langle f rangle$. Then $N_G(H) = langle f,r^4 rangle$ and $N_G( N_G(H) ) = langle f, r^2 rangle$ and $N_G( N_G( N_G(H) ) ) = langle f,r rangle =G$.
In other words, $H$ is not normal, neither is its normalizer, but the normalizer of the normalizer is normal.
Using dihedral groups of order $2^n$ you can create arbitrarily long chains of non-normal normalizers that eventually end up being normal.
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
$endgroup$
Let $G=langle f,r : f^2 = r^8 = 1, rf =fr^7 rangle$ be the dihedral group of order 16, and let $H=langle f rangle$. Then $N_G(H) = langle f,r^4 rangle$ and $N_G( N_G(H) ) = langle f, r^2 rangle$ and $N_G( N_G( N_G(H) ) ) = langle f,r rangle =G$.
In other words, $H$ is not normal, neither is its normalizer, but the normalizer of the normalizer is normal.
Using dihedral groups of order $2^n$ you can create arbitrarily long chains of non-normal normalizers that eventually end up being normal.
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
answered Apr 3 '14 at 15:58
Jack SchmidtJack Schmidt
43.2k572152
43.2k572152
add a comment |
add a comment |
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