Strongly p-embedded subgroups and p-Sylow subgroups.
Clash Royale CLAN TAG#URR8PPP
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Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap ^xH$ has order prime to $p,$ where $^xH$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:
Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap ^xH$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap ^yH $$ implies that $y in H.$ Thus $S subset H.$
Here is what I do not understand with the proof:
- Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap ^x H.$
- Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap ^xH$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap ^yH?$
group-theory
asked Sep 10 at 21:40
Dedalus
1,96211736
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1
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Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap ^xH$ has order prime to $p,$ where $^xH$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:
Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap ^xH$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap ^yH $$ implies that $y in H.$ Thus $S subset H.$
Here is what I do not understand with the proof:
- Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap ^x H.$
- Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap ^xH$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap ^yH?$
group-theory
asked Sep 10 at 21:40
Dedalus
1,96211736
I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
– James
Sep 10 at 22:28
@James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
– Dedalus
Sep 10 at 22:35
Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap ^yH$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
– James
Sep 10 at 22:45
Thanks, I misread the second statemnt and read it as $g in H cap ^yH.$
– Dedalus
Sep 10 at 22:55
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap ^xH$ has order prime to $p,$ where $^xH$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:
Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap ^xH$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap ^yH $$ implies that $y in H.$ Thus $S subset H.$
Here is what I do not understand with the proof:
- Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap ^x H.$
- Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap ^xH$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap ^yH?$
group-theory
asked Sep 10 at 21:40
Dedalus
1,96211736
Let $G$ be a finite group and $p$ a prime number. Recall that a subgroup $H$ of $G$ is strongly $p$-embedded if $p | |H|,$ and for each $x in G setminus H,$ $H cap ^xH$ has order prime to $p,$ where $^xH$ denotes the group we get after conjugation by $x.$ I am reading a proof which claims that any strongly $p$-embedded group contains a $p$-Sylow group of $G$. I can prove this fact by other means, but I would like to understand the proof I am reading. This is the argument I am reading:
Assume $H$ is strongly $p$-embedded. Since $p | |H|,$ $p$ contains an element $g$ of order $p.$ Choose any $S in Syl_p(G)$ such that $g in S.$ For $x in Z(S)$ of order $p,$ $$g in H cap ^xH$$ implies $x in H.$ Hence for all $y in S,$ $$x in H cap ^yH $$ implies that $y in H.$ Thus $S subset H.$
Here is what I do not understand with the proof:
- Why does it matter that $x in Z(S)$ has order $p?$ It seems to me that for any $x in Z(S),$ we have that $g in H cap ^x H.$
- Suppose I know that for $x in Z(S)$ of order $p,$ $g in H cap ^xH$ implies $x in H,$ i.e. that the elements of order $p$ in $Z(S)$ is a subset of $H.$ Why does it follow then that if $y in S$ is arbitrary, that $x in H cap ^yH?$
group-theory
group-theory
asked Sep 10 at 21:40
Dedalus
1,96211736
asked Sep 10 at 21:40
Dedalus
1,96211736
edited Sep 10 at 22:34
asked Sep 10 at 21:40
Dedalus
1,96211736
asked Sep 10 at 21:40
Dedalus
1,96211736
asked Sep 10 at 21:40
Dedalus
1,96211736
1,96211736
I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
– James
Sep 10 at 22:28
@James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
– Dedalus
Sep 10 at 22:35
Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap ^yH$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
– James
Sep 10 at 22:45
Thanks, I misread the second statemnt and read it as $g in H cap ^yH.$
– Dedalus
Sep 10 at 22:55
add a comment |Â
I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
– James
Sep 10 at 22:28
@James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
– Dedalus
Sep 10 at 22:35
Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap ^yH$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
– James
Sep 10 at 22:45
Thanks, I misread the second statemnt and read it as $g in H cap ^yH.$
– Dedalus
Sep 10 at 22:55
I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
– James
Sep 10 at 22:28
I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
– James
Sep 10 at 22:28
@James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
– Dedalus
Sep 10 at 22:35
@James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
– Dedalus
Sep 10 at 22:35
Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap ^yH$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
– James
Sep 10 at 22:45
Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap ^yH$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
– James
Sep 10 at 22:45
Thanks, I misread the second statemnt and read it as $g in H cap ^yH.$
– Dedalus
Sep 10 at 22:55
Thanks, I misread the second statemnt and read it as $g in H cap ^yH.$
– Dedalus
Sep 10 at 22:55
add a comment |Â
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I agree that for your first question the order of $x$ need not be equal to $p$. But the second question I find a bit confusing. Do you mean $Z(S)$ instead of $Z(P)$? If so, it seems to make sense.
– James
Sep 10 at 22:28
@James Thanks for pointing that out. Yes, I meant $Z(S).$ Does it make sense in the sense that you understand the argument of the proof, or in the sense that you understand my second question?
– Dedalus
Sep 10 at 22:35
Well, I now understand the question and, I think, also the argument. As per the first question, you need not restrict to elements of order $p$ so, in fact, we have $Z(S)leq H$ from the first part of the argument. And, there exists, therefore, $1neq tin Z(S)$ (so $t$ is also an element of $H$). So, if $y$ is now any element in $S$, we have $t = t^yin Hcap ^yH$ and has $p$-power order, so $y$ must belong to $H$ by the assumption that $H$ is strongly $p$-embedded. This gives $Sleq H$.
– James
Sep 10 at 22:45
Thanks, I misread the second statemnt and read it as $g in H cap ^yH.$
– Dedalus
Sep 10 at 22:55