Mixing
Sylow $p$-subgroups of Galois group
Clash Royale CLAN TAG#URR8PPP
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0
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It just my curious, but I couldn’t find any related concept:
Condition)
Let $G$ be a finite Galois group, and the number of each Sylow $p_i$-subgroup of $G$ is one, where $p_i$ is a prime factor of $|G|$.
Question)
Under the above condition, is it true $G$ is cyclic?
Give some advice or related notion! Thank you!
abstract-algebra galois-theory
edited Jul 27 at 8:53
Bernard
110k635102
asked Jul 27 at 5:23
Primavera
1548
add a comment |Â
up vote
0
down vote
favorite
It just my curious, but I couldn’t find any related concept:
Condition)
Let $G$ be a finite Galois group, and the number of each Sylow $p_i$-subgroup of $G$ is one, where $p_i$ is a prime factor of $|G|$.
Question)
Under the above condition, is it true $G$ is cyclic?
Give some advice or related notion! Thank you!
abstract-algebra galois-theory
edited Jul 27 at 8:53
Bernard
110k635102
asked Jul 27 at 5:23
Primavera
1548
3
All Sylow subgroups being normal is well known to be equivalent to the group $G$ being nilpotent. There are non-cyclic nilpotent groups. In fact every $p$-group is nilpotent ...
– Jyrki Lahtonen
Jul 27 at 5:33
@JyrkiLahtonen Actually, I was solving the following problems: Let $K$ be a Galois extension over $mathbbQ$ with $[K:mathbbQ]=540$, let $G:=G(K/mathbbQ)$ be the Galois group so that the number of each Sylow subgroup of $G$ is one. I don't know who made it, the solution of the above problem claims that $G$ is cyclic without any proof. I don't understand how to guarantee $G$ is cyclic. Since $Gcong P_2times P_3times P_5$, where each $P_i$ is a Sylow $i$-subgroup of $G$, I think there is no guarantee that $G$ is cyclic. Is there something I missed?
– Primavera
Jul 27 at 14:32
$540=2^2cdot3^3cdot5$. There are non-cyclic groups of orders $2^2$ and $3^3$, so you cannot conclude that $G$ would be cyclic based on that piece of data alone. Yes, something is missing from that exercise.
– Jyrki Lahtonen
Jul 27 at 14:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It just my curious, but I couldn’t find any related concept:
Condition)
Let $G$ be a finite Galois group, and the number of each Sylow $p_i$-subgroup of $G$ is one, where $p_i$ is a prime factor of $|G|$.
Question)
Under the above condition, is it true $G$ is cyclic?
Give some advice or related notion! Thank you!
abstract-algebra galois-theory
edited Jul 27 at 8:53
Bernard
110k635102
asked Jul 27 at 5:23
Primavera
1548
It just my curious, but I couldn’t find any related concept:
Condition)
Let $G$ be a finite Galois group, and the number of each Sylow $p_i$-subgroup of $G$ is one, where $p_i$ is a prime factor of $|G|$.
Question)
Under the above condition, is it true $G$ is cyclic?
Give some advice or related notion! Thank you!
abstract-algebra galois-theory
edited Jul 27 at 8:53
Bernard
110k635102
asked Jul 27 at 5:23
Primavera
1548
edited Jul 27 at 8:53
Bernard
110k635102
edited Jul 27 at 8:53
Bernard
110k635102
edited Jul 27 at 8:53
Bernard
110k635102
110k635102
asked Jul 27 at 5:23
Primavera
1548
asked Jul 27 at 5:23
Primavera
1548
asked Jul 27 at 5:23
Primavera
1548
1548
3
All Sylow subgroups being normal is well known to be equivalent to the group $G$ being nilpotent. There are non-cyclic nilpotent groups. In fact every $p$-group is nilpotent ...
– Jyrki Lahtonen
Jul 27 at 5:33
@JyrkiLahtonen Actually, I was solving the following problems: Let $K$ be a Galois extension over $mathbbQ$ with $[K:mathbbQ]=540$, let $G:=G(K/mathbbQ)$ be the Galois group so that the number of each Sylow subgroup of $G$ is one. I don't know who made it, the solution of the above problem claims that $G$ is cyclic without any proof. I don't understand how to guarantee $G$ is cyclic. Since $Gcong P_2times P_3times P_5$, where each $P_i$ is a Sylow $i$-subgroup of $G$, I think there is no guarantee that $G$ is cyclic. Is there something I missed?
– Primavera
Jul 27 at 14:32
$540=2^2cdot3^3cdot5$. There are non-cyclic groups of orders $2^2$ and $3^3$, so you cannot conclude that $G$ would be cyclic based on that piece of data alone. Yes, something is missing from that exercise.
– Jyrki Lahtonen
Jul 27 at 14:48
add a comment |Â
3
All Sylow subgroups being normal is well known to be equivalent to the group $G$ being nilpotent. There are non-cyclic nilpotent groups. In fact every $p$-group is nilpotent ...
– Jyrki Lahtonen
Jul 27 at 5:33
@JyrkiLahtonen Actually, I was solving the following problems: Let $K$ be a Galois extension over $mathbbQ$ with $[K:mathbbQ]=540$, let $G:=G(K/mathbbQ)$ be the Galois group so that the number of each Sylow subgroup of $G$ is one. I don't know who made it, the solution of the above problem claims that $G$ is cyclic without any proof. I don't understand how to guarantee $G$ is cyclic. Since $Gcong P_2times P_3times P_5$, where each $P_i$ is a Sylow $i$-subgroup of $G$, I think there is no guarantee that $G$ is cyclic. Is there something I missed?
– Primavera
Jul 27 at 14:32
$540=2^2cdot3^3cdot5$. There are non-cyclic groups of orders $2^2$ and $3^3$, so you cannot conclude that $G$ would be cyclic based on that piece of data alone. Yes, something is missing from that exercise.
– Jyrki Lahtonen
Jul 27 at 14:48
3
3
All Sylow subgroups being normal is well known to be equivalent to the group $G$ being nilpotent. There are non-cyclic nilpotent groups. In fact every $p$-group is nilpotent ...
– Jyrki Lahtonen
Jul 27 at 5:33
All Sylow subgroups being normal is well known to be equivalent to the group $G$ being nilpotent. There are non-cyclic nilpotent groups. In fact every $p$-group is nilpotent ...
– Jyrki Lahtonen
Jul 27 at 5:33
@JyrkiLahtonen Actually, I was solving the following problems: Let $K$ be a Galois extension over $mathbbQ$ with $[K:mathbbQ]=540$, let $G:=G(K/mathbbQ)$ be the Galois group so that the number of each Sylow subgroup of $G$ is one. I don't know who made it, the solution of the above problem claims that $G$ is cyclic without any proof. I don't understand how to guarantee $G$ is cyclic. Since $Gcong P_2times P_3times P_5$, where each $P_i$ is a Sylow $i$-subgroup of $G$, I think there is no guarantee that $G$ is cyclic. Is there something I missed?
– Primavera
Jul 27 at 14:32
@JyrkiLahtonen Actually, I was solving the following problems: Let $K$ be a Galois extension over $mathbbQ$ with $[K:mathbbQ]=540$, let $G:=G(K/mathbbQ)$ be the Galois group so that the number of each Sylow subgroup of $G$ is one. I don't know who made it, the solution of the above problem claims that $G$ is cyclic without any proof. I don't understand how to guarantee $G$ is cyclic. Since $Gcong P_2times P_3times P_5$, where each $P_i$ is a Sylow $i$-subgroup of $G$, I think there is no guarantee that $G$ is cyclic. Is there something I missed?
– Primavera
Jul 27 at 14:32
$540=2^2cdot3^3cdot5$. There are non-cyclic groups of orders $2^2$ and $3^3$, so you cannot conclude that $G$ would be cyclic based on that piece of data alone. Yes, something is missing from that exercise.
– Jyrki Lahtonen
Jul 27 at 14:48
$540=2^2cdot3^3cdot5$. There are non-cyclic groups of orders $2^2$ and $3^3$, so you cannot conclude that $G$ would be cyclic based on that piece of data alone. Yes, something is missing from that exercise.
– Jyrki Lahtonen
Jul 27 at 14:48
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
Every finite group is a Galois group of some finite extension $L/K$
of fields.
If a finite group has a unique Sylow subgroup for each prime, then all
one can conclude is that it is nilpotent, that is a direct
product of $p$-groups. It need not be cyclic, or indeed Abelian.
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
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
Every finite group is a Galois group of some finite extension $L/K$
of fields.
If a finite group has a unique Sylow subgroup for each prime, then all
one can conclude is that it is nilpotent, that is a direct
product of $p$-groups. It need not be cyclic, or indeed Abelian.
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
add a comment |Â
up vote
4
down vote
Every finite group is a Galois group of some finite extension $L/K$
of fields.
If a finite group has a unique Sylow subgroup for each prime, then all
one can conclude is that it is nilpotent, that is a direct
product of $p$-groups. It need not be cyclic, or indeed Abelian.
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Every finite group is a Galois group of some finite extension $L/K$
of fields.
If a finite group has a unique Sylow subgroup for each prime, then all
one can conclude is that it is nilpotent, that is a direct
product of $p$-groups. It need not be cyclic, or indeed Abelian.
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
Every finite group is a Galois group of some finite extension $L/K$
of fields.
If a finite group has a unique Sylow subgroup for each prime, then all
one can conclude is that it is nilpotent, that is a direct
product of $p$-groups. It need not be cyclic, or indeed Abelian.
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
answered Jul 27 at 5:33
Lord Shark the Unknown
84.6k950111
84.6k950111
add a comment |Â
add a comment |Â
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3
All Sylow subgroups being normal is well known to be equivalent to the group $G$ being nilpotent. There are non-cyclic nilpotent groups. In fact every $p$-group is nilpotent ...
– Jyrki Lahtonen
Jul 27 at 5:33
@JyrkiLahtonen Actually, I was solving the following problems: Let $K$ be a Galois extension over $mathbbQ$ with $[K:mathbbQ]=540$, let $G:=G(K/mathbbQ)$ be the Galois group so that the number of each Sylow subgroup of $G$ is one. I don't know who made it, the solution of the above problem claims that $G$ is cyclic without any proof. I don't understand how to guarantee $G$ is cyclic. Since $Gcong P_2times P_3times P_5$, where each $P_i$ is a Sylow $i$-subgroup of $G$, I think there is no guarantee that $G$ is cyclic. Is there something I missed?
– Primavera
Jul 27 at 14:32
$540=2^2cdot3^3cdot5$. There are non-cyclic groups of orders $2^2$ and $3^3$, so you cannot conclude that $G$ would be cyclic based on that piece of data alone. Yes, something is missing from that exercise.
– Jyrki Lahtonen
Jul 27 at 14:48