Proof of Sylow's second and third theorem from Lang's book
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This is the excerpt of the proof of Sylow's theorem from Lang's and some moments of the are unclear to me.
1) In order to prove (ii) they take $H$ to be Sylow $p$-subgroup and it is not obvious to me why any two Sylow $p$-subgroups are conjugate?
2) Is $H$ is a Sylow $p$-subgroup then why it has only one fixed point?
I have spent some hours in order to understand these questions but was not able.
Would be very grateful for detailed help!
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
asked Nov 4 at 22:11
RFZ
4,83931337
add a comment |
up vote
1
down vote
favorite
This is the excerpt of the proof of Sylow's theorem from Lang's and some moments of the are unclear to me.
1) In order to prove (ii) they take $H$ to be Sylow $p$-subgroup and it is not obvious to me why any two Sylow $p$-subgroups are conjugate?
2) Is $H$ is a Sylow $p$-subgroup then why it has only one fixed point?
I have spent some hours in order to understand these questions but was not able.
Would be very grateful for detailed help!
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
asked Nov 4 at 22:11
RFZ
4,83931337
1
It's not clear to me what you don't understand in your first question. Do you not understand why $H=Q$ in the first sentence you underlined? Or is it just that you don't understand how to deduce (ii) from $H=Q$?
– Eric Wofsey
Nov 4 at 22:16
@EricWofsey, Sorry for my not well-organized question. Let me clarify it: We want to prove (ii) via (i). Let's take $H$ to be Sylow $p$-subgroup then $H$ is contained in some another Sylow $p$-subgroup, call it $Q$. So we have $Hsubset Q$ $Rightarrow$ $H=Q$ because their order are equal. How it follow from here that any two Sylow $p$-subgroups are conjugate?
– RFZ
Nov 4 at 22:22
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is the excerpt of the proof of Sylow's theorem from Lang's and some moments of the are unclear to me.
1) In order to prove (ii) they take $H$ to be Sylow $p$-subgroup and it is not obvious to me why any two Sylow $p$-subgroups are conjugate?
2) Is $H$ is a Sylow $p$-subgroup then why it has only one fixed point?
I have spent some hours in order to understand these questions but was not able.
Would be very grateful for detailed help!
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
asked Nov 4 at 22:11
RFZ
4,83931337
This is the excerpt of the proof of Sylow's theorem from Lang's and some moments of the are unclear to me.
1) In order to prove (ii) they take $H$ to be Sylow $p$-subgroup and it is not obvious to me why any two Sylow $p$-subgroups are conjugate?
2) Is $H$ is a Sylow $p$-subgroup then why it has only one fixed point?
I have spent some hours in order to understand these questions but was not able.
Would be very grateful for detailed help!
abstract-algebra group-theory finite-groups sylow-theory
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
asked Nov 4 at 22:11
RFZ
4,83931337
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
asked Nov 4 at 22:11
RFZ
4,83931337
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
edited Nov 4 at 22:25
Eric Wofsey
173k12201324
173k12201324
asked Nov 4 at 22:11
RFZ
4,83931337
asked Nov 4 at 22:11
RFZ
4,83931337
asked Nov 4 at 22:11
RFZ
4,83931337
4,83931337
1
It's not clear to me what you don't understand in your first question. Do you not understand why $H=Q$ in the first sentence you underlined? Or is it just that you don't understand how to deduce (ii) from $H=Q$?
– Eric Wofsey
Nov 4 at 22:16
@EricWofsey, Sorry for my not well-organized question. Let me clarify it: We want to prove (ii) via (i). Let's take $H$ to be Sylow $p$-subgroup then $H$ is contained in some another Sylow $p$-subgroup, call it $Q$. So we have $Hsubset Q$ $Rightarrow$ $H=Q$ because their order are equal. How it follow from here that any two Sylow $p$-subgroups are conjugate?
– RFZ
Nov 4 at 22:22
add a comment |
1
It's not clear to me what you don't understand in your first question. Do you not understand why $H=Q$ in the first sentence you underlined? Or is it just that you don't understand how to deduce (ii) from $H=Q$?
– Eric Wofsey
Nov 4 at 22:16
@EricWofsey, Sorry for my not well-organized question. Let me clarify it: We want to prove (ii) via (i). Let's take $H$ to be Sylow $p$-subgroup then $H$ is contained in some another Sylow $p$-subgroup, call it $Q$. So we have $Hsubset Q$ $Rightarrow$ $H=Q$ because their order are equal. How it follow from here that any two Sylow $p$-subgroups are conjugate?
– RFZ
Nov 4 at 22:22
1
1
It's not clear to me what you don't understand in your first question. Do you not understand why $H=Q$ in the first sentence you underlined? Or is it just that you don't understand how to deduce (ii) from $H=Q$?
– Eric Wofsey
Nov 4 at 22:16
It's not clear to me what you don't understand in your first question. Do you not understand why $H=Q$ in the first sentence you underlined? Or is it just that you don't understand how to deduce (ii) from $H=Q$?
– Eric Wofsey
Nov 4 at 22:16
@EricWofsey, Sorry for my not well-organized question. Let me clarify it: We want to prove (ii) via (i). Let's take $H$ to be Sylow $p$-subgroup then $H$ is contained in some another Sylow $p$-subgroup, call it $Q$. So we have $Hsubset Q$ $Rightarrow$ $H=Q$ because their order are equal. How it follow from here that any two Sylow $p$-subgroups are conjugate?
– RFZ
Nov 4 at 22:22
@EricWofsey, Sorry for my not well-organized question. Let me clarify it: We want to prove (ii) via (i). Let's take $H$ to be Sylow $p$-subgroup then $H$ is contained in some another Sylow $p$-subgroup, call it $Q$. So we have $Hsubset Q$ $Rightarrow$ $H=Q$ because their order are equal. How it follow from here that any two Sylow $p$-subgroups are conjugate?
– RFZ
Nov 4 at 22:22
add a comment |
1 Answer
1
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oldest
votes
up vote
4
down vote
accepted
The argument shows that for any $p$-Sylow subgroup $H$, there exists $Qin S$ such that $H=Q$. But $S$ was by definition the set of conjugates of $P$, so this means $H$ is conjugate to $P$. So, this proves that every $p$-Sylow subgroup $H$ is conjugate to $P$, which proves (ii).
Moreover, the argument shows that if $H$ is $p$-Sylow and $Q$ is any fixed point of the action of $H$ on $S$, then $H=Q$. So, there can only be one such fixed point, namely $H$.
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
1
Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1
– RFZ
Nov 4 at 22:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The argument shows that for any $p$-Sylow subgroup $H$, there exists $Qin S$ such that $H=Q$. But $S$ was by definition the set of conjugates of $P$, so this means $H$ is conjugate to $P$. So, this proves that every $p$-Sylow subgroup $H$ is conjugate to $P$, which proves (ii).
Moreover, the argument shows that if $H$ is $p$-Sylow and $Q$ is any fixed point of the action of $H$ on $S$, then $H=Q$. So, there can only be one such fixed point, namely $H$.
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
1
Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1
– RFZ
Nov 4 at 22:31
add a comment |
up vote
4
down vote
accepted
The argument shows that for any $p$-Sylow subgroup $H$, there exists $Qin S$ such that $H=Q$. But $S$ was by definition the set of conjugates of $P$, so this means $H$ is conjugate to $P$. So, this proves that every $p$-Sylow subgroup $H$ is conjugate to $P$, which proves (ii).
Moreover, the argument shows that if $H$ is $p$-Sylow and $Q$ is any fixed point of the action of $H$ on $S$, then $H=Q$. So, there can only be one such fixed point, namely $H$.
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
1
Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1
– RFZ
Nov 4 at 22:31
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The argument shows that for any $p$-Sylow subgroup $H$, there exists $Qin S$ such that $H=Q$. But $S$ was by definition the set of conjugates of $P$, so this means $H$ is conjugate to $P$. So, this proves that every $p$-Sylow subgroup $H$ is conjugate to $P$, which proves (ii).
Moreover, the argument shows that if $H$ is $p$-Sylow and $Q$ is any fixed point of the action of $H$ on $S$, then $H=Q$. So, there can only be one such fixed point, namely $H$.
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
The argument shows that for any $p$-Sylow subgroup $H$, there exists $Qin S$ such that $H=Q$. But $S$ was by definition the set of conjugates of $P$, so this means $H$ is conjugate to $P$. So, this proves that every $p$-Sylow subgroup $H$ is conjugate to $P$, which proves (ii).
Moreover, the argument shows that if $H$ is $p$-Sylow and $Q$ is any fixed point of the action of $H$ on $S$, then $H=Q$. So, there can only be one such fixed point, namely $H$.
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
answered Nov 4 at 22:24
Eric Wofsey
173k12201324
173k12201324
1
Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1
– RFZ
Nov 4 at 22:31
add a comment |
1
Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1
– RFZ
Nov 4 at 22:31
1
1
Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1
– RFZ
Nov 4 at 22:31
Dear Eric, right now with your help i have totally got it! Thank you very much for your help! I was thinking in the wrong direction. +1
– RFZ
Nov 4 at 22:31
add a comment |
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1
It's not clear to me what you don't understand in your first question. Do you not understand why $H=Q$ in the first sentence you underlined? Or is it just that you don't understand how to deduce (ii) from $H=Q$?
– Eric Wofsey
Nov 4 at 22:16
@EricWofsey, Sorry for my not well-organized question. Let me clarify it: We want to prove (ii) via (i). Let's take $H$ to be Sylow $p$-subgroup then $H$ is contained in some another Sylow $p$-subgroup, call it $Q$. So we have $Hsubset Q$ $Rightarrow$ $H=Q$ because their order are equal. How it follow from here that any two Sylow $p$-subgroups are conjugate?
– RFZ
Nov 4 at 22:22