Small step in a Sylow theorem proof Announcing the arrival of Valued Associate #679: Cesar...
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Small step in a Sylow theorem proof
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)A group of order 30 has a normal 5-Sylow subgroup.Sylow theorem Cyclic sylow p subgroupSylow p- subgroupsLang Sylow TheoremProof of Sylow by PinterAbout first Sylow Theorem proofNormalizers of subgroups of Sylow $p$-subgroupsIs the normalizer of a Sylow $p$-subgroup is a $p$-group?Question on Sylow subgroup and normalizerIn a group of order 400, is the normalizer of one of the 16 Sylow 5-subgroups itself?
$begingroup$
I'm following
https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowmore.pdf
and in particular the following paragraph:
Let $N = N_G(P)$ be the normalizer of $P$ in $G$. Then all the elements of p-power order in N lie in P. Indeed, any element of N with p-power order which is not in P would give a
non-identity element of p-power order in N/P. Then we could take inverse images through the projection N → N/P to find a p-subgroup inside N properly containing P, but this contradicts the maximality of P as a p-subgroup of G.
My question in the proof by contradiction.
$|z|=p^k$, $zin N$ but $znotin P$, then $nPne P$. Why is $nP$ an element of $p$-power? $(nP)^j=n^jP$, and thus $n^j=e$ only if $j$ is a p-power, but what about $n^jin P$ with $j$ not p-power? Then we would have $|nP|=j$.
$pi:Nto N/P$, since $nPne P$, $pi^{-1}(nP)$ cannot be a group since $e$ does not belong to it. What is meant in the text? Perhaps $langle pi^{-1}(nP)rangle$?
abstract-algebra sylow-theory
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
$endgroup$
|
show 2 more comments
$begingroup$
I'm following
https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowmore.pdf
and in particular the following paragraph:
Let $N = N_G(P)$ be the normalizer of $P$ in $G$. Then all the elements of p-power order in N lie in P. Indeed, any element of N with p-power order which is not in P would give a
non-identity element of p-power order in N/P. Then we could take inverse images through the projection N → N/P to find a p-subgroup inside N properly containing P, but this contradicts the maximality of P as a p-subgroup of G.
My question in the proof by contradiction.
$|z|=p^k$, $zin N$ but $znotin P$, then $nPne P$. Why is $nP$ an element of $p$-power? $(nP)^j=n^jP$, and thus $n^j=e$ only if $j$ is a p-power, but what about $n^jin P$ with $j$ not p-power? Then we would have $|nP|=j$.
$pi:Nto N/P$, since $nPne P$, $pi^{-1}(nP)$ cannot be a group since $e$ does not belong to it. What is meant in the text? Perhaps $langle pi^{-1}(nP)rangle$?
abstract-algebra sylow-theory
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
$endgroup$
$begingroup$
Hint for (1): $x^k=1$ implies $(xH)^k=H$; so in general $o(xH)|o(x)$.
$endgroup$
– ancientmathematician
Mar 25 at 9:02
$begingroup$
Your $z$ should be $n$.
$endgroup$
– KCd
Mar 25 at 10:51
$begingroup$
I updated the file to add another explanation of that point.
$endgroup$
– KCd
Mar 25 at 14:16
$begingroup$
@KCd Thank you very much. The only other unclear point was in the last paragraph of section 2: I couldn't figure out what "fixed point congruence" is? I used the orbit-stabilizer theorem instead. Since $N$ is the only fixed point of the action of $P$ on $G/N$ by left multiplication, then the orbit of $N$, $Gcdot N$, is the only $1$ element orbit; the order of the other orbits is larger than one and divides $|P|$ (O-S theorem), hence the other orbit have a $p$-power order, hence $|G/N|=[G:N]=1+mp$ or $[G:N]equiv 1 pmod p$.[...]
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
[...] Concluding (since $Ptriangleleft Ntriangleleft G$) $[G:P]=[G:N][N:P]equiv [N:P]notequiv 0pmod{p}$, where we have used the previously proven fact that $pnmid [N:P]$. The equality shows that the order of $P$ is largest $p$-order in $|G|$.
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
|
show 2 more comments
$begingroup$
I'm following
https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowmore.pdf
and in particular the following paragraph:
Let $N = N_G(P)$ be the normalizer of $P$ in $G$. Then all the elements of p-power order in N lie in P. Indeed, any element of N with p-power order which is not in P would give a
non-identity element of p-power order in N/P. Then we could take inverse images through the projection N → N/P to find a p-subgroup inside N properly containing P, but this contradicts the maximality of P as a p-subgroup of G.
My question in the proof by contradiction.
$|z|=p^k$, $zin N$ but $znotin P$, then $nPne P$. Why is $nP$ an element of $p$-power? $(nP)^j=n^jP$, and thus $n^j=e$ only if $j$ is a p-power, but what about $n^jin P$ with $j$ not p-power? Then we would have $|nP|=j$.
$pi:Nto N/P$, since $nPne P$, $pi^{-1}(nP)$ cannot be a group since $e$ does not belong to it. What is meant in the text? Perhaps $langle pi^{-1}(nP)rangle$?
abstract-algebra sylow-theory
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
$endgroup$
I'm following
https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowmore.pdf
and in particular the following paragraph:
Let $N = N_G(P)$ be the normalizer of $P$ in $G$. Then all the elements of p-power order in N lie in P. Indeed, any element of N with p-power order which is not in P would give a
non-identity element of p-power order in N/P. Then we could take inverse images through the projection N → N/P to find a p-subgroup inside N properly containing P, but this contradicts the maximality of P as a p-subgroup of G.
My question in the proof by contradiction.
$|z|=p^k$, $zin N$ but $znotin P$, then $nPne P$. Why is $nP$ an element of $p$-power? $(nP)^j=n^jP$, and thus $n^j=e$ only if $j$ is a p-power, but what about $n^jin P$ with $j$ not p-power? Then we would have $|nP|=j$.
$pi:Nto N/P$, since $nPne P$, $pi^{-1}(nP)$ cannot be a group since $e$ does not belong to it. What is meant in the text? Perhaps $langle pi^{-1}(nP)rangle$?
abstract-algebra sylow-theory
abstract-algebra sylow-theory
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
asked Mar 25 at 8:47
Aaron LenzAaron Lenz
7110
7110
$begingroup$
Hint for (1): $x^k=1$ implies $(xH)^k=H$; so in general $o(xH)|o(x)$.
$endgroup$
– ancientmathematician
Mar 25 at 9:02
$begingroup$
Your $z$ should be $n$.
$endgroup$
– KCd
Mar 25 at 10:51
$begingroup$
I updated the file to add another explanation of that point.
$endgroup$
– KCd
Mar 25 at 14:16
$begingroup$
@KCd Thank you very much. The only other unclear point was in the last paragraph of section 2: I couldn't figure out what "fixed point congruence" is? I used the orbit-stabilizer theorem instead. Since $N$ is the only fixed point of the action of $P$ on $G/N$ by left multiplication, then the orbit of $N$, $Gcdot N$, is the only $1$ element orbit; the order of the other orbits is larger than one and divides $|P|$ (O-S theorem), hence the other orbit have a $p$-power order, hence $|G/N|=[G:N]=1+mp$ or $[G:N]equiv 1 pmod p$.[...]
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
[...] Concluding (since $Ptriangleleft Ntriangleleft G$) $[G:P]=[G:N][N:P]equiv [N:P]notequiv 0pmod{p}$, where we have used the previously proven fact that $pnmid [N:P]$. The equality shows that the order of $P$ is largest $p$-order in $|G|$.
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
|
show 2 more comments
$begingroup$
Hint for (1): $x^k=1$ implies $(xH)^k=H$; so in general $o(xH)|o(x)$.
$endgroup$
– ancientmathematician
Mar 25 at 9:02
$begingroup$
Your $z$ should be $n$.
$endgroup$
– KCd
Mar 25 at 10:51
$begingroup$
I updated the file to add another explanation of that point.
$endgroup$
– KCd
Mar 25 at 14:16
$begingroup$
@KCd Thank you very much. The only other unclear point was in the last paragraph of section 2: I couldn't figure out what "fixed point congruence" is? I used the orbit-stabilizer theorem instead. Since $N$ is the only fixed point of the action of $P$ on $G/N$ by left multiplication, then the orbit of $N$, $Gcdot N$, is the only $1$ element orbit; the order of the other orbits is larger than one and divides $|P|$ (O-S theorem), hence the other orbit have a $p$-power order, hence $|G/N|=[G:N]=1+mp$ or $[G:N]equiv 1 pmod p$.[...]
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
[...] Concluding (since $Ptriangleleft Ntriangleleft G$) $[G:P]=[G:N][N:P]equiv [N:P]notequiv 0pmod{p}$, where we have used the previously proven fact that $pnmid [N:P]$. The equality shows that the order of $P$ is largest $p$-order in $|G|$.
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
Hint for (1): $x^k=1$ implies $(xH)^k=H$; so in general $o(xH)|o(x)$.
$endgroup$
– ancientmathematician
Mar 25 at 9:02
$begingroup$
Hint for (1): $x^k=1$ implies $(xH)^k=H$; so in general $o(xH)|o(x)$.
$endgroup$
– ancientmathematician
Mar 25 at 9:02
$begingroup$
Your $z$ should be $n$.
$endgroup$
– KCd
Mar 25 at 10:51
$begingroup$
Your $z$ should be $n$.
$endgroup$
– KCd
Mar 25 at 10:51
$begingroup$
I updated the file to add another explanation of that point.
$endgroup$
– KCd
Mar 25 at 14:16
$begingroup$
I updated the file to add another explanation of that point.
$endgroup$
– KCd
Mar 25 at 14:16
$begingroup$
@KCd Thank you very much. The only other unclear point was in the last paragraph of section 2: I couldn't figure out what "fixed point congruence" is? I used the orbit-stabilizer theorem instead. Since $N$ is the only fixed point of the action of $P$ on $G/N$ by left multiplication, then the orbit of $N$, $Gcdot N$, is the only $1$ element orbit; the order of the other orbits is larger than one and divides $|P|$ (O-S theorem), hence the other orbit have a $p$-power order, hence $|G/N|=[G:N]=1+mp$ or $[G:N]equiv 1 pmod p$.[...]
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
@KCd Thank you very much. The only other unclear point was in the last paragraph of section 2: I couldn't figure out what "fixed point congruence" is? I used the orbit-stabilizer theorem instead. Since $N$ is the only fixed point of the action of $P$ on $G/N$ by left multiplication, then the orbit of $N$, $Gcdot N$, is the only $1$ element orbit; the order of the other orbits is larger than one and divides $|P|$ (O-S theorem), hence the other orbit have a $p$-power order, hence $|G/N|=[G:N]=1+mp$ or $[G:N]equiv 1 pmod p$.[...]
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
[...] Concluding (since $Ptriangleleft Ntriangleleft G$) $[G:P]=[G:N][N:P]equiv [N:P]notequiv 0pmod{p}$, where we have used the previously proven fact that $pnmid [N:P]$. The equality shows that the order of $P$ is largest $p$-order in $|G|$.
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
[...] Concluding (since $Ptriangleleft Ntriangleleft G$) $[G:P]=[G:N][N:P]equiv [N:P]notequiv 0pmod{p}$, where we have used the previously proven fact that $pnmid [N:P]$. The equality shows that the order of $P$ is largest $p$-order in $|G|$.
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The answer to 2. is "Yes", what is meant is $langle pi^{-1}(nP)rangle$.
For 1, I find it easiest to think about $pi$ as just 'any' homomorphism from $N$ to 'some group' $Q$ (so we forget everything we know about $Q$ and $pi$ except that $pi$ is a homomorphism.) The claim now is that if $n in N$ has $p$-power order in $N$, then so does $pi(n)$ in $Q$.
In fact it is even worse: the order of $pi(n)$ divides the order of $n$.
It all comes down to Euclid's algorithm from elementary number theory. Let $a$ be the order of $n$ in $N$ and $b$ be the order of $pi(n)$ in $Q$. Using 'division with remainder' we have that $a = qb + r$ for some natural numbers $q$ and $r$ with $r < b$ (possibly $r = 0$).
Let $e$ be the unit of $Q$. Then $e = pi(n^a) =pi((n^b)^q n^r) = (pi(n)^b)^q pi(n)^r = e^q pi(n)^r = e pi(n)^r = pi(n)^r$, or shorter: $pi(n)^r = e$.
But since $b$ was the smallest positive number such that $pi(n)^b = e$, and $r < b$ we have that $r = 0$ and hence $a = qb$ so that indeed $b|a$.
answered Mar 25 at 9:02
VincentVincent
3,33811231
$endgroup$
add a comment |
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$begingroup$
The answer to 2. is "Yes", what is meant is $langle pi^{-1}(nP)rangle$.
For 1, I find it easiest to think about $pi$ as just 'any' homomorphism from $N$ to 'some group' $Q$ (so we forget everything we know about $Q$ and $pi$ except that $pi$ is a homomorphism.) The claim now is that if $n in N$ has $p$-power order in $N$, then so does $pi(n)$ in $Q$.
In fact it is even worse: the order of $pi(n)$ divides the order of $n$.
It all comes down to Euclid's algorithm from elementary number theory. Let $a$ be the order of $n$ in $N$ and $b$ be the order of $pi(n)$ in $Q$. Using 'division with remainder' we have that $a = qb + r$ for some natural numbers $q$ and $r$ with $r < b$ (possibly $r = 0$).
Let $e$ be the unit of $Q$. Then $e = pi(n^a) =pi((n^b)^q n^r) = (pi(n)^b)^q pi(n)^r = e^q pi(n)^r = e pi(n)^r = pi(n)^r$, or shorter: $pi(n)^r = e$.
But since $b$ was the smallest positive number such that $pi(n)^b = e$, and $r < b$ we have that $r = 0$ and hence $a = qb$ so that indeed $b|a$.
answered Mar 25 at 9:02
VincentVincent
3,33811231
$endgroup$
add a comment |
$begingroup$
The answer to 2. is "Yes", what is meant is $langle pi^{-1}(nP)rangle$.
For 1, I find it easiest to think about $pi$ as just 'any' homomorphism from $N$ to 'some group' $Q$ (so we forget everything we know about $Q$ and $pi$ except that $pi$ is a homomorphism.) The claim now is that if $n in N$ has $p$-power order in $N$, then so does $pi(n)$ in $Q$.
In fact it is even worse: the order of $pi(n)$ divides the order of $n$.
It all comes down to Euclid's algorithm from elementary number theory. Let $a$ be the order of $n$ in $N$ and $b$ be the order of $pi(n)$ in $Q$. Using 'division with remainder' we have that $a = qb + r$ for some natural numbers $q$ and $r$ with $r < b$ (possibly $r = 0$).
Let $e$ be the unit of $Q$. Then $e = pi(n^a) =pi((n^b)^q n^r) = (pi(n)^b)^q pi(n)^r = e^q pi(n)^r = e pi(n)^r = pi(n)^r$, or shorter: $pi(n)^r = e$.
But since $b$ was the smallest positive number such that $pi(n)^b = e$, and $r < b$ we have that $r = 0$ and hence $a = qb$ so that indeed $b|a$.
answered Mar 25 at 9:02
VincentVincent
3,33811231
$endgroup$
add a comment |
$begingroup$
The answer to 2. is "Yes", what is meant is $langle pi^{-1}(nP)rangle$.
For 1, I find it easiest to think about $pi$ as just 'any' homomorphism from $N$ to 'some group' $Q$ (so we forget everything we know about $Q$ and $pi$ except that $pi$ is a homomorphism.) The claim now is that if $n in N$ has $p$-power order in $N$, then so does $pi(n)$ in $Q$.
In fact it is even worse: the order of $pi(n)$ divides the order of $n$.
It all comes down to Euclid's algorithm from elementary number theory. Let $a$ be the order of $n$ in $N$ and $b$ be the order of $pi(n)$ in $Q$. Using 'division with remainder' we have that $a = qb + r$ for some natural numbers $q$ and $r$ with $r < b$ (possibly $r = 0$).
Let $e$ be the unit of $Q$. Then $e = pi(n^a) =pi((n^b)^q n^r) = (pi(n)^b)^q pi(n)^r = e^q pi(n)^r = e pi(n)^r = pi(n)^r$, or shorter: $pi(n)^r = e$.
But since $b$ was the smallest positive number such that $pi(n)^b = e$, and $r < b$ we have that $r = 0$ and hence $a = qb$ so that indeed $b|a$.
answered Mar 25 at 9:02
VincentVincent
3,33811231
$endgroup$
The answer to 2. is "Yes", what is meant is $langle pi^{-1}(nP)rangle$.
For 1, I find it easiest to think about $pi$ as just 'any' homomorphism from $N$ to 'some group' $Q$ (so we forget everything we know about $Q$ and $pi$ except that $pi$ is a homomorphism.) The claim now is that if $n in N$ has $p$-power order in $N$, then so does $pi(n)$ in $Q$.
In fact it is even worse: the order of $pi(n)$ divides the order of $n$.
It all comes down to Euclid's algorithm from elementary number theory. Let $a$ be the order of $n$ in $N$ and $b$ be the order of $pi(n)$ in $Q$. Using 'division with remainder' we have that $a = qb + r$ for some natural numbers $q$ and $r$ with $r < b$ (possibly $r = 0$).
Let $e$ be the unit of $Q$. Then $e = pi(n^a) =pi((n^b)^q n^r) = (pi(n)^b)^q pi(n)^r = e^q pi(n)^r = e pi(n)^r = pi(n)^r$, or shorter: $pi(n)^r = e$.
But since $b$ was the smallest positive number such that $pi(n)^b = e$, and $r < b$ we have that $r = 0$ and hence $a = qb$ so that indeed $b|a$.
answered Mar 25 at 9:02
VincentVincent
3,33811231
answered Mar 25 at 9:02
VincentVincent
3,33811231
answered Mar 25 at 9:02
VincentVincent
3,33811231
answered Mar 25 at 9:02
VincentVincent
3,33811231
3,33811231
add a comment |
add a comment |
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$begingroup$
Hint for (1): $x^k=1$ implies $(xH)^k=H$; so in general $o(xH)|o(x)$.
$endgroup$
– ancientmathematician
Mar 25 at 9:02
$begingroup$
Your $z$ should be $n$.
$endgroup$
– KCd
Mar 25 at 10:51
$begingroup$
I updated the file to add another explanation of that point.
$endgroup$
– KCd
Mar 25 at 14:16
$begingroup$
@KCd Thank you very much. The only other unclear point was in the last paragraph of section 2: I couldn't figure out what "fixed point congruence" is? I used the orbit-stabilizer theorem instead. Since $N$ is the only fixed point of the action of $P$ on $G/N$ by left multiplication, then the orbit of $N$, $Gcdot N$, is the only $1$ element orbit; the order of the other orbits is larger than one and divides $|P|$ (O-S theorem), hence the other orbit have a $p$-power order, hence $|G/N|=[G:N]=1+mp$ or $[G:N]equiv 1 pmod p$.[...]
$endgroup$
– Aaron Lenz
Mar 26 at 7:32
$begingroup$
[...] Concluding (since $Ptriangleleft Ntriangleleft G$) $[G:P]=[G:N][N:P]equiv [N:P]notequiv 0pmod{p}$, where we have used the previously proven fact that $pnmid [N:P]$. The equality shows that the order of $P$ is largest $p$-order in $|G|$.
$endgroup$
– Aaron Lenz
Mar 26 at 7:32