If $Ntrianglelefteq G$, then $phi(N) leq phi(G)$, where $phi(N)$ is the Frattini subgroup of $N$.
$begingroup$
I was thinking somehow to use normality of N as follows
Since N is normal, then $G/N$ will be a group, so we can consider the
natural map
$pi : G rightarrow G/N$, where $g mapsto gN$, $ker(pi) = N$.
So It is here enough to prove that a maximal subgroup M of G will contain
$ker(pi)$, then since $phi(G)$ is intersection all such maximal subgroup M, so we will get that $phi(N) leq phi(G)$, but I got stuck in this specific step.
abstract-algebra
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
$endgroup$
add a comment |
$begingroup$
I was thinking somehow to use normality of N as follows
Since N is normal, then $G/N$ will be a group, so we can consider the
natural map
$pi : G rightarrow G/N$, where $g mapsto gN$, $ker(pi) = N$.
So It is here enough to prove that a maximal subgroup M of G will contain
$ker(pi)$, then since $phi(G)$ is intersection all such maximal subgroup M, so we will get that $phi(N) leq phi(G)$, but I got stuck in this specific step.
abstract-algebra
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
$endgroup$
add a comment |
$begingroup$
I was thinking somehow to use normality of N as follows
Since N is normal, then $G/N$ will be a group, so we can consider the
natural map
$pi : G rightarrow G/N$, where $g mapsto gN$, $ker(pi) = N$.
So It is here enough to prove that a maximal subgroup M of G will contain
$ker(pi)$, then since $phi(G)$ is intersection all such maximal subgroup M, so we will get that $phi(N) leq phi(G)$, but I got stuck in this specific step.
abstract-algebra
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
$endgroup$
I was thinking somehow to use normality of N as follows
Since N is normal, then $G/N$ will be a group, so we can consider the
natural map
$pi : G rightarrow G/N$, where $g mapsto gN$, $ker(pi) = N$.
So It is here enough to prove that a maximal subgroup M of G will contain
$ker(pi)$, then since $phi(G)$ is intersection all such maximal subgroup M, so we will get that $phi(N) leq phi(G)$, but I got stuck in this specific step.
abstract-algebra
abstract-algebra
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
edited Jan 3 at 20:09
Dietrich Burde
81.5k648106
81.5k648106
asked Oct 12 '15 at 6:46
user111750
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$begingroup$
The statement has been discussed here:
For Frattini subgroup $Phi(G)$, if $N vartriangleleft G$, then $Phi(N) leq Phi(G)$. : Counter-example
We have the following result (for finite groups it coincides with the above statement):
Theorem: The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.
Proof: The Frattini subgroup $Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $Phi(N)$ is a normal subgroup of G.
We need to show that $Phi(N)$ is contained in $Phi(G)$. For this, it suffices to show that $Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction.
Suppose $M$ is a maximal subgroup of $G$ not containing $Phi(N)$. Then, the subgroup generated by $M$ and $Phi(N)$ is equal to $G$. Since $Phi(N)$ is normal in $G$, we have $MPhi(N) = G$. From that, it follows that
$Phi(N)(M cap N) = N$.
Since $M$ does not contain $Phi(N)$, $M$ does not contain $N$ either and hence
$M cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M cap N$ and $Phi(N)$. This contradicts the fact that their product is $N$.
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
add a comment |
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$begingroup$
The statement has been discussed here:
For Frattini subgroup $Phi(G)$, if $N vartriangleleft G$, then $Phi(N) leq Phi(G)$. : Counter-example
We have the following result (for finite groups it coincides with the above statement):
Theorem: The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.
Proof: The Frattini subgroup $Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $Phi(N)$ is a normal subgroup of G.
We need to show that $Phi(N)$ is contained in $Phi(G)$. For this, it suffices to show that $Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction.
Suppose $M$ is a maximal subgroup of $G$ not containing $Phi(N)$. Then, the subgroup generated by $M$ and $Phi(N)$ is equal to $G$. Since $Phi(N)$ is normal in $G$, we have $MPhi(N) = G$. From that, it follows that
$Phi(N)(M cap N) = N$.
Since $M$ does not contain $Phi(N)$, $M$ does not contain $N$ either and hence
$M cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M cap N$ and $Phi(N)$. This contradicts the fact that their product is $N$.
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
add a comment |
$begingroup$
The statement has been discussed here:
For Frattini subgroup $Phi(G)$, if $N vartriangleleft G$, then $Phi(N) leq Phi(G)$. : Counter-example
We have the following result (for finite groups it coincides with the above statement):
Theorem: The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.
Proof: The Frattini subgroup $Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $Phi(N)$ is a normal subgroup of G.
We need to show that $Phi(N)$ is contained in $Phi(G)$. For this, it suffices to show that $Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction.
Suppose $M$ is a maximal subgroup of $G$ not containing $Phi(N)$. Then, the subgroup generated by $M$ and $Phi(N)$ is equal to $G$. Since $Phi(N)$ is normal in $G$, we have $MPhi(N) = G$. From that, it follows that
$Phi(N)(M cap N) = N$.
Since $M$ does not contain $Phi(N)$, $M$ does not contain $N$ either and hence
$M cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M cap N$ and $Phi(N)$. This contradicts the fact that their product is $N$.
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
add a comment |
$begingroup$
The statement has been discussed here:
For Frattini subgroup $Phi(G)$, if $N vartriangleleft G$, then $Phi(N) leq Phi(G)$. : Counter-example
We have the following result (for finite groups it coincides with the above statement):
Theorem: The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.
Proof: The Frattini subgroup $Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $Phi(N)$ is a normal subgroup of G.
We need to show that $Phi(N)$ is contained in $Phi(G)$. For this, it suffices to show that $Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction.
Suppose $M$ is a maximal subgroup of $G$ not containing $Phi(N)$. Then, the subgroup generated by $M$ and $Phi(N)$ is equal to $G$. Since $Phi(N)$ is normal in $G$, we have $MPhi(N) = G$. From that, it follows that
$Phi(N)(M cap N) = N$.
Since $M$ does not contain $Phi(N)$, $M$ does not contain $N$ either and hence
$M cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M cap N$ and $Phi(N)$. This contradicts the fact that their product is $N$.
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
The statement has been discussed here:
For Frattini subgroup $Phi(G)$, if $N vartriangleleft G$, then $Phi(N) leq Phi(G)$. : Counter-example
We have the following result (for finite groups it coincides with the above statement):
Theorem: The Frattini subgroup of any normal subgroup is contained in the Frattini subgroup of the whole group, provided the normal subgroup is a group in which every proper subgroup is contained in a maximal subgroup.
Proof: The Frattini subgroup $Phi(N)$ is a characteristic subgroup of $N$. Since every characteristic subgroup of a normal subgroup is normal, $Phi(N)$ is a normal subgroup of G.
We need to show that $Phi(N)$ is contained in $Phi(G)$. For this, it suffices to show that $Phi(N)$ is contained in every maximal subgroup of $G$. We prove this by contradiction.
Suppose $M$ is a maximal subgroup of $G$ not containing $Phi(N)$. Then, the subgroup generated by $M$ and $Phi(N)$ is equal to $G$. Since $Phi(N)$ is normal in $G$, we have $MPhi(N) = G$. From that, it follows that
$Phi(N)(M cap N) = N$.
Since $M$ does not contain $Phi(N)$, $M$ does not contain $N$ either and hence
$M cap N$ is a proper subgroup of $N$. Since every proper subgroup is contained in a maximal subgroup of $N$, there is a maximal subgroup of $N$ containing both $M cap N$ and $Phi(N)$. This contradicts the fact that their product is $N$.
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
answered Jan 3 at 20:15
Dietrich BurdeDietrich Burde
81.5k648106
81.5k648106
add a comment |
add a comment |
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