Proposition 1.39, Hatcher - Quotient By A Non-Normal Subgroup
This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.
general-topology algebraic-topology
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
add a comment |
This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.
general-topology algebraic-topology
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
1
But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04
Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06
No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07
add a comment |
This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.
general-topology algebraic-topology
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.
general-topology algebraic-topology
general-topology algebraic-topology
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
asked Nov 27 '18 at 2:02
Frederic Chopin
321111
321111
1
But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04
Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06
No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07
add a comment |
1
But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04
Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06
No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07
1
1
But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04
But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04
Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06
Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06
No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07
No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07
add a comment |
1 Answer
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(I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
add a comment |
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(I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
add a comment |
(I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
add a comment |
(I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
(I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
answered Nov 27 '18 at 2:21
Antonios-Alexandros Robotis
9,20741640
9,20741640
add a comment |
add a comment |
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1
But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04
Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06
No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07