Mixing
$P$ is a $p$-Sylow subgroup of $G$ and $H$ is a subgroup of $G$. Show that $Pcap H$ is contained in a...
$begingroup$
Let $G$ a group, let $H$ a subgroup of $G$ and let $Pinoperatorname{Syl}_p(G)$. Show that $Pcap H$ is a subset of some $Uinoperatorname{Syl}_p(H)$.
I know that begin{align} |G| &=p^acdot nk,\ |H| &=p^bcdot n, \ |P| &=p^aend{align} for $bleq a, (p,n),(p,k)=1$. The trivial case is $a=b$, but i don't know how to handle the case $b<a$.
abstract-algebra group-theory finite-groups
edited Jan 22 at 20:35
Shaun
9,381113684
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
$endgroup$
add a comment |
$begingroup$
Let $G$ a group, let $H$ a subgroup of $G$ and let $Pinoperatorname{Syl}_p(G)$. Show that $Pcap H$ is a subset of some $Uinoperatorname{Syl}_p(H)$.
I know that begin{align} |G| &=p^acdot nk,\ |H| &=p^bcdot n, \ |P| &=p^aend{align} for $bleq a, (p,n),(p,k)=1$. The trivial case is $a=b$, but i don't know how to handle the case $b<a$.
abstract-algebra group-theory finite-groups
edited Jan 22 at 20:35
Shaun
9,381113684
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
$endgroup$
$begingroup$
@J.Doe.It's Sylow and it is pronounced as "See-loe"
$endgroup$
– Chinnapparaj R
Jan 21 at 11:27
$begingroup$
@DavidMitra If $|G|=p^acdot m$ for some prime $p, 0leq a$ and $(p,m)=1$ then p-Silow subgroups family denoted by $operatorname{Syl}_p(G)$ is the family of all the subgroups of $G$ of order $p^a$.
$endgroup$
– J. Doe
Jan 21 at 11:28
2
$begingroup$
@ChinnapparajR Technically it's pronounced "see-lov".
$endgroup$
– Yuval Gat
Jan 21 at 11:52
$begingroup$
@YuvalGat: Yes! agreed!
$endgroup$
– Chinnapparaj R
Jan 21 at 11:57
$begingroup$
A fun generalistion-of-a-special-case that, despite being an algebraist, I found so implausible that I disbelieved it when someone told me: if $P$ is normal in $G$, then $H cap P$ is a Sylow $p$-subgroup of $H$.
$endgroup$
– LSpice
Jan 22 at 21:03
add a comment |
$begingroup$
Let $G$ a group, let $H$ a subgroup of $G$ and let $Pinoperatorname{Syl}_p(G)$. Show that $Pcap H$ is a subset of some $Uinoperatorname{Syl}_p(H)$.
I know that begin{align} |G| &=p^acdot nk,\ |H| &=p^bcdot n, \ |P| &=p^aend{align} for $bleq a, (p,n),(p,k)=1$. The trivial case is $a=b$, but i don't know how to handle the case $b<a$.
abstract-algebra group-theory finite-groups
edited Jan 22 at 20:35
Shaun
9,381113684
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
$endgroup$
Let $G$ a group, let $H$ a subgroup of $G$ and let $Pinoperatorname{Syl}_p(G)$. Show that $Pcap H$ is a subset of some $Uinoperatorname{Syl}_p(H)$.
I know that begin{align} |G| &=p^acdot nk,\ |H| &=p^bcdot n, \ |P| &=p^aend{align} for $bleq a, (p,n),(p,k)=1$. The trivial case is $a=b$, but i don't know how to handle the case $b<a$.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Jan 22 at 20:35
Shaun
9,381113684
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
edited Jan 22 at 20:35
Shaun
9,381113684
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
edited Jan 22 at 20:35
Shaun
9,381113684
edited Jan 22 at 20:35
Shaun
9,381113684
edited Jan 22 at 20:35
Shaun
9,381113684
9,381113684
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
asked Jan 21 at 11:24
J. DoeJ. Doe
16611
16611
$begingroup$
@J.Doe.It's Sylow and it is pronounced as "See-loe"
$endgroup$
– Chinnapparaj R
Jan 21 at 11:27
$begingroup$
@DavidMitra If $|G|=p^acdot m$ for some prime $p, 0leq a$ and $(p,m)=1$ then p-Silow subgroups family denoted by $operatorname{Syl}_p(G)$ is the family of all the subgroups of $G$ of order $p^a$.
$endgroup$
– J. Doe
Jan 21 at 11:28
2
$begingroup$
@ChinnapparajR Technically it's pronounced "see-lov".
$endgroup$
– Yuval Gat
Jan 21 at 11:52
$begingroup$
@YuvalGat: Yes! agreed!
$endgroup$
– Chinnapparaj R
Jan 21 at 11:57
$begingroup$
A fun generalistion-of-a-special-case that, despite being an algebraist, I found so implausible that I disbelieved it when someone told me: if $P$ is normal in $G$, then $H cap P$ is a Sylow $p$-subgroup of $H$.
$endgroup$
– LSpice
Jan 22 at 21:03
add a comment |
$begingroup$
@J.Doe.It's Sylow and it is pronounced as "See-loe"
$endgroup$
– Chinnapparaj R
Jan 21 at 11:27
$begingroup$
@DavidMitra If $|G|=p^acdot m$ for some prime $p, 0leq a$ and $(p,m)=1$ then p-Silow subgroups family denoted by $operatorname{Syl}_p(G)$ is the family of all the subgroups of $G$ of order $p^a$.
$endgroup$
– J. Doe
Jan 21 at 11:28
2
$begingroup$
@ChinnapparajR Technically it's pronounced "see-lov".
$endgroup$
– Yuval Gat
Jan 21 at 11:52
$begingroup$
@YuvalGat: Yes! agreed!
$endgroup$
– Chinnapparaj R
Jan 21 at 11:57
$begingroup$
A fun generalistion-of-a-special-case that, despite being an algebraist, I found so implausible that I disbelieved it when someone told me: if $P$ is normal in $G$, then $H cap P$ is a Sylow $p$-subgroup of $H$.
$endgroup$
– LSpice
Jan 22 at 21:03
$begingroup$
@J.Doe.It's Sylow and it is pronounced as "See-loe"
$endgroup$
– Chinnapparaj R
Jan 21 at 11:27
$begingroup$
@J.Doe.It's Sylow and it is pronounced as "See-loe"
$endgroup$
– Chinnapparaj R
Jan 21 at 11:27
$begingroup$
@DavidMitra If $|G|=p^acdot m$ for some prime $p, 0leq a$ and $(p,m)=1$ then p-Silow subgroups family denoted by $operatorname{Syl}_p(G)$ is the family of all the subgroups of $G$ of order $p^a$.
$endgroup$
– J. Doe
Jan 21 at 11:28
$begingroup$
@DavidMitra If $|G|=p^acdot m$ for some prime $p, 0leq a$ and $(p,m)=1$ then p-Silow subgroups family denoted by $operatorname{Syl}_p(G)$ is the family of all the subgroups of $G$ of order $p^a$.
$endgroup$
– J. Doe
Jan 21 at 11:28
2
2
$begingroup$
@ChinnapparajR Technically it's pronounced "see-lov".
$endgroup$
– Yuval Gat
Jan 21 at 11:52
$begingroup$
@ChinnapparajR Technically it's pronounced "see-lov".
$endgroup$
– Yuval Gat
Jan 21 at 11:52
$begingroup$
@YuvalGat: Yes! agreed!
$endgroup$
– Chinnapparaj R
Jan 21 at 11:57
$begingroup$
@YuvalGat: Yes! agreed!
$endgroup$
– Chinnapparaj R
Jan 21 at 11:57
$begingroup$
A fun generalistion-of-a-special-case that, despite being an algebraist, I found so implausible that I disbelieved it when someone told me: if $P$ is normal in $G$, then $H cap P$ is a Sylow $p$-subgroup of $H$.
$endgroup$
– LSpice
Jan 22 at 21:03
$begingroup$
A fun generalistion-of-a-special-case that, despite being an algebraist, I found so implausible that I disbelieved it when someone told me: if $P$ is normal in $G$, then $H cap P$ is a Sylow $p$-subgroup of $H$.
$endgroup$
– LSpice
Jan 22 at 21:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is a little hint:
what's the order of the elements of $P cap H $ ?
Then use Sylow's theorem to deduce that this is contain in a P-sylow of H.
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
$endgroup$
$begingroup$
I think it may make more sense just to ask what is the order of $P cap H$ itself, to avoid the application of Cauchy's theorem that looking at the order of its elements would require.
$endgroup$
– LSpice
Jan 22 at 20:37
$begingroup$
Don't need Cauchy's Theorem. A P-Sylow is in particular a p-group. So all the elements in $ P cap H $ have order a power of $ p $. It's a p-group, then you can apply Sylow's theorem.
$endgroup$
– Cauchy is my master
Jan 22 at 20:47
$begingroup$
Yes, but you're stating the converse of what is needed. Your argument will show that all the elements of $P cap H$ have order a $p$-power; but then one needs Cauchy's theorem to show that $P cap H$ is a $p$-group (in the sense that its order is a $p$-power), which is easier to establish directly because the order of $P cap H$ divides that of $P$.
$endgroup$
– LSpice
Jan 22 at 20:53
1
$begingroup$
Oh sorry, I just misread what you wrote. You are absolutely right, Lagrange theorem better applies here.
$endgroup$
– Cauchy is my master
Jan 22 at 20:56
add a comment |
$begingroup$
Being the intersection of two subgroups, $P cap H$ is a subgroup. Moreover, it is contained in $P$, so it has order $p^k$ for some $k$.
Let $Q$ be a maximal $p$-subgroup of $H$ that contains $P cap H$. It has to exist since you can take $Q_0 = P cap H$, and then search for a bigger $Q_1$. If you can't find it, then $Q=Q_0$. Otherwise, take $Q_1$ and search for a bigger $Q_2$, and so on. This process has to terminate, since $H$ is finite.
The key is that now, from Sylow's theorems, we know that every maximal $p$-subgroup of $H$ is a Sylow $p$-subgroup of $H$. So take $U=Q$ and you are done.
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a little hint:
what's the order of the elements of $P cap H $ ?
Then use Sylow's theorem to deduce that this is contain in a P-sylow of H.
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
$endgroup$
$begingroup$
I think it may make more sense just to ask what is the order of $P cap H$ itself, to avoid the application of Cauchy's theorem that looking at the order of its elements would require.
$endgroup$
– LSpice
Jan 22 at 20:37
$begingroup$
Don't need Cauchy's Theorem. A P-Sylow is in particular a p-group. So all the elements in $ P cap H $ have order a power of $ p $. It's a p-group, then you can apply Sylow's theorem.
$endgroup$
– Cauchy is my master
Jan 22 at 20:47
$begingroup$
Yes, but you're stating the converse of what is needed. Your argument will show that all the elements of $P cap H$ have order a $p$-power; but then one needs Cauchy's theorem to show that $P cap H$ is a $p$-group (in the sense that its order is a $p$-power), which is easier to establish directly because the order of $P cap H$ divides that of $P$.
$endgroup$
– LSpice
Jan 22 at 20:53
1
$begingroup$
Oh sorry, I just misread what you wrote. You are absolutely right, Lagrange theorem better applies here.
$endgroup$
– Cauchy is my master
Jan 22 at 20:56
add a comment |
$begingroup$
Here is a little hint:
what's the order of the elements of $P cap H $ ?
Then use Sylow's theorem to deduce that this is contain in a P-sylow of H.
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
$endgroup$
$begingroup$
I think it may make more sense just to ask what is the order of $P cap H$ itself, to avoid the application of Cauchy's theorem that looking at the order of its elements would require.
$endgroup$
– LSpice
Jan 22 at 20:37
$begingroup$
Don't need Cauchy's Theorem. A P-Sylow is in particular a p-group. So all the elements in $ P cap H $ have order a power of $ p $. It's a p-group, then you can apply Sylow's theorem.
$endgroup$
– Cauchy is my master
Jan 22 at 20:47
$begingroup$
Yes, but you're stating the converse of what is needed. Your argument will show that all the elements of $P cap H$ have order a $p$-power; but then one needs Cauchy's theorem to show that $P cap H$ is a $p$-group (in the sense that its order is a $p$-power), which is easier to establish directly because the order of $P cap H$ divides that of $P$.
$endgroup$
– LSpice
Jan 22 at 20:53
1
$begingroup$
Oh sorry, I just misread what you wrote. You are absolutely right, Lagrange theorem better applies here.
$endgroup$
– Cauchy is my master
Jan 22 at 20:56
add a comment |
$begingroup$
Here is a little hint:
what's the order of the elements of $P cap H $ ?
Then use Sylow's theorem to deduce that this is contain in a P-sylow of H.
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
$endgroup$
Here is a little hint:
what's the order of the elements of $P cap H $ ?
Then use Sylow's theorem to deduce that this is contain in a P-sylow of H.
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
edited Jan 21 at 11:32
Chinnapparaj R
5,6932928
5,6932928
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
answered Jan 21 at 11:31
Cauchy is my masterCauchy is my master
413
413
$begingroup$
I think it may make more sense just to ask what is the order of $P cap H$ itself, to avoid the application of Cauchy's theorem that looking at the order of its elements would require.
$endgroup$
– LSpice
Jan 22 at 20:37
$begingroup$
Don't need Cauchy's Theorem. A P-Sylow is in particular a p-group. So all the elements in $ P cap H $ have order a power of $ p $. It's a p-group, then you can apply Sylow's theorem.
$endgroup$
– Cauchy is my master
Jan 22 at 20:47
$begingroup$
Yes, but you're stating the converse of what is needed. Your argument will show that all the elements of $P cap H$ have order a $p$-power; but then one needs Cauchy's theorem to show that $P cap H$ is a $p$-group (in the sense that its order is a $p$-power), which is easier to establish directly because the order of $P cap H$ divides that of $P$.
$endgroup$
– LSpice
Jan 22 at 20:53
1
$begingroup$
Oh sorry, I just misread what you wrote. You are absolutely right, Lagrange theorem better applies here.
$endgroup$
– Cauchy is my master
Jan 22 at 20:56
add a comment |
$begingroup$
I think it may make more sense just to ask what is the order of $P cap H$ itself, to avoid the application of Cauchy's theorem that looking at the order of its elements would require.
$endgroup$
– LSpice
Jan 22 at 20:37
$begingroup$
Don't need Cauchy's Theorem. A P-Sylow is in particular a p-group. So all the elements in $ P cap H $ have order a power of $ p $. It's a p-group, then you can apply Sylow's theorem.
$endgroup$
– Cauchy is my master
Jan 22 at 20:47
$begingroup$
Yes, but you're stating the converse of what is needed. Your argument will show that all the elements of $P cap H$ have order a $p$-power; but then one needs Cauchy's theorem to show that $P cap H$ is a $p$-group (in the sense that its order is a $p$-power), which is easier to establish directly because the order of $P cap H$ divides that of $P$.
$endgroup$
– LSpice
Jan 22 at 20:53
1
$begingroup$
Oh sorry, I just misread what you wrote. You are absolutely right, Lagrange theorem better applies here.
$endgroup$
– Cauchy is my master
Jan 22 at 20:56
$begingroup$
I think it may make more sense just to ask what is the order of $P cap H$ itself, to avoid the application of Cauchy's theorem that looking at the order of its elements would require.
$endgroup$
– LSpice
Jan 22 at 20:37
$begingroup$
I think it may make more sense just to ask what is the order of $P cap H$ itself, to avoid the application of Cauchy's theorem that looking at the order of its elements would require.
$endgroup$
– LSpice
Jan 22 at 20:37
$begingroup$
Don't need Cauchy's Theorem. A P-Sylow is in particular a p-group. So all the elements in $ P cap H $ have order a power of $ p $. It's a p-group, then you can apply Sylow's theorem.
$endgroup$
– Cauchy is my master
Jan 22 at 20:47
$begingroup$
Don't need Cauchy's Theorem. A P-Sylow is in particular a p-group. So all the elements in $ P cap H $ have order a power of $ p $. It's a p-group, then you can apply Sylow's theorem.
$endgroup$
– Cauchy is my master
Jan 22 at 20:47
$begingroup$
Yes, but you're stating the converse of what is needed. Your argument will show that all the elements of $P cap H$ have order a $p$-power; but then one needs Cauchy's theorem to show that $P cap H$ is a $p$-group (in the sense that its order is a $p$-power), which is easier to establish directly because the order of $P cap H$ divides that of $P$.
$endgroup$
– LSpice
Jan 22 at 20:53
$begingroup$
Yes, but you're stating the converse of what is needed. Your argument will show that all the elements of $P cap H$ have order a $p$-power; but then one needs Cauchy's theorem to show that $P cap H$ is a $p$-group (in the sense that its order is a $p$-power), which is easier to establish directly because the order of $P cap H$ divides that of $P$.
$endgroup$
– LSpice
Jan 22 at 20:53
1
1
$begingroup$
Oh sorry, I just misread what you wrote. You are absolutely right, Lagrange theorem better applies here.
$endgroup$
– Cauchy is my master
Jan 22 at 20:56
$begingroup$
Oh sorry, I just misread what you wrote. You are absolutely right, Lagrange theorem better applies here.
$endgroup$
– Cauchy is my master
Jan 22 at 20:56
add a comment |
$begingroup$
Being the intersection of two subgroups, $P cap H$ is a subgroup. Moreover, it is contained in $P$, so it has order $p^k$ for some $k$.
Let $Q$ be a maximal $p$-subgroup of $H$ that contains $P cap H$. It has to exist since you can take $Q_0 = P cap H$, and then search for a bigger $Q_1$. If you can't find it, then $Q=Q_0$. Otherwise, take $Q_1$ and search for a bigger $Q_2$, and so on. This process has to terminate, since $H$ is finite.
The key is that now, from Sylow's theorems, we know that every maximal $p$-subgroup of $H$ is a Sylow $p$-subgroup of $H$. So take $U=Q$ and you are done.
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
$endgroup$
add a comment |
$begingroup$
Being the intersection of two subgroups, $P cap H$ is a subgroup. Moreover, it is contained in $P$, so it has order $p^k$ for some $k$.
Let $Q$ be a maximal $p$-subgroup of $H$ that contains $P cap H$. It has to exist since you can take $Q_0 = P cap H$, and then search for a bigger $Q_1$. If you can't find it, then $Q=Q_0$. Otherwise, take $Q_1$ and search for a bigger $Q_2$, and so on. This process has to terminate, since $H$ is finite.
The key is that now, from Sylow's theorems, we know that every maximal $p$-subgroup of $H$ is a Sylow $p$-subgroup of $H$. So take $U=Q$ and you are done.
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
$endgroup$
add a comment |
$begingroup$
Being the intersection of two subgroups, $P cap H$ is a subgroup. Moreover, it is contained in $P$, so it has order $p^k$ for some $k$.
Let $Q$ be a maximal $p$-subgroup of $H$ that contains $P cap H$. It has to exist since you can take $Q_0 = P cap H$, and then search for a bigger $Q_1$. If you can't find it, then $Q=Q_0$. Otherwise, take $Q_1$ and search for a bigger $Q_2$, and so on. This process has to terminate, since $H$ is finite.
The key is that now, from Sylow's theorems, we know that every maximal $p$-subgroup of $H$ is a Sylow $p$-subgroup of $H$. So take $U=Q$ and you are done.
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
$endgroup$
Being the intersection of two subgroups, $P cap H$ is a subgroup. Moreover, it is contained in $P$, so it has order $p^k$ for some $k$.
Let $Q$ be a maximal $p$-subgroup of $H$ that contains $P cap H$. It has to exist since you can take $Q_0 = P cap H$, and then search for a bigger $Q_1$. If you can't find it, then $Q=Q_0$. Otherwise, take $Q_1$ and search for a bigger $Q_2$, and so on. This process has to terminate, since $H$ is finite.
The key is that now, from Sylow's theorems, we know that every maximal $p$-subgroup of $H$ is a Sylow $p$-subgroup of $H$. So take $U=Q$ and you are done.
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
answered Jan 21 at 11:37
AnalysisStudent0414AnalysisStudent0414
4,389928
4,389928
add a comment |
add a comment |
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$begingroup$
@J.Doe.It's Sylow and it is pronounced as "See-loe"
$endgroup$
– Chinnapparaj R
Jan 21 at 11:27
$begingroup$
@DavidMitra If $|G|=p^acdot m$ for some prime $p, 0leq a$ and $(p,m)=1$ then p-Silow subgroups family denoted by $operatorname{Syl}_p(G)$ is the family of all the subgroups of $G$ of order $p^a$.
$endgroup$
– J. Doe
Jan 21 at 11:28
2
$begingroup$
@ChinnapparajR Technically it's pronounced "see-lov".
$endgroup$
– Yuval Gat
Jan 21 at 11:52
$begingroup$
@YuvalGat: Yes! agreed!
$endgroup$
– Chinnapparaj R
Jan 21 at 11:57
$begingroup$
A fun generalistion-of-a-special-case that, despite being an algebraist, I found so implausible that I disbelieved it when someone told me: if $P$ is normal in $G$, then $H cap P$ is a Sylow $p$-subgroup of $H$.
$endgroup$
– LSpice
Jan 22 at 21:03