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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 3 hours ago
Eric Wofsey
173k12199321
asked 4 hours ago
RFZ
4,79731337
add a comment |Â
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 3 hours ago
Eric Wofsey
173k12199321
asked 4 hours ago
RFZ
4,79731337
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
3 hours ago
@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
3 hours ago
add a comment |Â
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 3 hours ago
Eric Wofsey
173k12199321
asked 4 hours ago
RFZ
4,79731337
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 3 hours ago
Eric Wofsey
173k12199321
asked 4 hours ago
RFZ
4,79731337
edited 3 hours ago
Eric Wofsey
173k12199321
asked 4 hours ago
RFZ
4,79731337
edited 3 hours ago
Eric Wofsey
173k12199321
edited 3 hours ago
Eric Wofsey
173k12199321
edited 3 hours ago
Eric Wofsey
173k12199321
173k12199321
asked 4 hours ago
RFZ
4,79731337
asked 4 hours ago
RFZ
4,79731337
asked 4 hours ago
RFZ
4,79731337
4,79731337
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
3 hours ago
@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
3 hours ago
add a comment |Â
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
3 hours ago
@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
3 hours ago
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
3 hours ago
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
3 hours ago
@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
3 hours ago
@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
3 hours ago
add a comment |Â
1 Answer
1
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 3 hours ago
Eric Wofsey
173k12199321
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
3 hours ago
add a comment |Â
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 3 hours ago
Eric Wofsey
173k12199321
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
3 hours ago
add a comment |Â
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 3 hours ago
Eric Wofsey
173k12199321
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
3 hours ago
add a comment |Â
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 3 hours ago
Eric Wofsey
173k12199321
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 3 hours ago
Eric Wofsey
173k12199321
answered 3 hours ago
Eric Wofsey
173k12199321
answered 3 hours ago
Eric Wofsey
173k12199321
answered 3 hours ago
Eric Wofsey
173k12199321
173k12199321
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
3 hours ago
add a comment |Â
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
3 hours ago
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
3 hours ago
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
3 hours ago
add a comment |Â
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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
3 hours ago
@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
3 hours ago