Mixing
Frattini subgroup of a subgroup
$begingroup$
I am doing an exercise for Frattini subgroups:
Let $G$ be a finite group and $N$ a normal subgroup of $G$. Show that there exists a subgroup $K$ of $G$ with $G=KN$ and $K cap N leq Phi(K)$.
I have no idea what theorem/result would give the existence of such $K$? This $K$ looks similar to a complement to $N$ but I don't think it is indeed the complement. Or maybe I have to construct such a group?
Any hints/comments are appreciated! Thanks!
group-theory finite-groups frattini-subgroup
edited Jan 28 at 19:22
the_fox
2,90031538
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
$endgroup$
add a comment |
$begingroup$
I am doing an exercise for Frattini subgroups:
Let $G$ be a finite group and $N$ a normal subgroup of $G$. Show that there exists a subgroup $K$ of $G$ with $G=KN$ and $K cap N leq Phi(K)$.
I have no idea what theorem/result would give the existence of such $K$? This $K$ looks similar to a complement to $N$ but I don't think it is indeed the complement. Or maybe I have to construct such a group?
Any hints/comments are appreciated! Thanks!
group-theory finite-groups frattini-subgroup
edited Jan 28 at 19:22
the_fox
2,90031538
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
$endgroup$
add a comment |
$begingroup$
I am doing an exercise for Frattini subgroups:
Let $G$ be a finite group and $N$ a normal subgroup of $G$. Show that there exists a subgroup $K$ of $G$ with $G=KN$ and $K cap N leq Phi(K)$.
I have no idea what theorem/result would give the existence of such $K$? This $K$ looks similar to a complement to $N$ but I don't think it is indeed the complement. Or maybe I have to construct such a group?
Any hints/comments are appreciated! Thanks!
group-theory finite-groups frattini-subgroup
edited Jan 28 at 19:22
the_fox
2,90031538
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
$endgroup$
I am doing an exercise for Frattini subgroups:
Let $G$ be a finite group and $N$ a normal subgroup of $G$. Show that there exists a subgroup $K$ of $G$ with $G=KN$ and $K cap N leq Phi(K)$.
I have no idea what theorem/result would give the existence of such $K$? This $K$ looks similar to a complement to $N$ but I don't think it is indeed the complement. Or maybe I have to construct such a group?
Any hints/comments are appreciated! Thanks!
group-theory finite-groups frattini-subgroup
group-theory finite-groups frattini-subgroup
edited Jan 28 at 19:22
the_fox
2,90031538
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
edited Jan 28 at 19:22
the_fox
2,90031538
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
edited Jan 28 at 19:22
the_fox
2,90031538
edited Jan 28 at 19:22
the_fox
2,90031538
edited Jan 28 at 19:22
the_fox
2,90031538
2,90031538
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
asked Oct 31 '18 at 21:54
Zheng XiaoZheng Xiao
37719
37719
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the set of supplements to $N$ in $G$. That is, let
$$mathcal{X} = {H leq G : HN = G}.$$
Clearly $mathcal{X}$ is non-empty because $G in mathcal{X}$. Now let $K$ be a minimal element of $mathcal{X}$, in the sense that $K$ does not contain properly any other element from $mathcal{X}$. By definition of $mathcal{X}$ we have $KN = G$. Now let $M$ be a maximal subgroup of $K$. We argue that $K cap N leq M$. Once we have proved that, it will follow that every maximal subgroup of $K$ contains $K cap N$, thus $K cap N leq Phi(K)$.
Suppose for a contradiction that $K cap N$ is not a subgroup of $M$. Since $N$ is normal in $G$, $K cap N$ is normal in $K$. So $K cap N$ permutes with every subgroup of $K$, thus also with $M$. So $(K cap N)M$ is a subgroup of $K$ which contains $M$ properly. Therefore $K = (K cap N)M$. Now notice that
$$G = KN = (K cap N)MN = (K cap N)NM = NM = MN,$$
so $M$ supplements $N$ in $G$, against the minimal choice of $K$.
We see, therefore, that $K cap N$ is contained in every maximal subgroup of $K$ and so $K cap N leq Phi(K)$, as wanted.
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
$endgroup$
1
$begingroup$
Thank you very much! Your proof is easy to understand.
$endgroup$
– Zheng Xiao
Nov 1 '18 at 3:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider the set of supplements to $N$ in $G$. That is, let
$$mathcal{X} = {H leq G : HN = G}.$$
Clearly $mathcal{X}$ is non-empty because $G in mathcal{X}$. Now let $K$ be a minimal element of $mathcal{X}$, in the sense that $K$ does not contain properly any other element from $mathcal{X}$. By definition of $mathcal{X}$ we have $KN = G$. Now let $M$ be a maximal subgroup of $K$. We argue that $K cap N leq M$. Once we have proved that, it will follow that every maximal subgroup of $K$ contains $K cap N$, thus $K cap N leq Phi(K)$.
Suppose for a contradiction that $K cap N$ is not a subgroup of $M$. Since $N$ is normal in $G$, $K cap N$ is normal in $K$. So $K cap N$ permutes with every subgroup of $K$, thus also with $M$. So $(K cap N)M$ is a subgroup of $K$ which contains $M$ properly. Therefore $K = (K cap N)M$. Now notice that
$$G = KN = (K cap N)MN = (K cap N)NM = NM = MN,$$
so $M$ supplements $N$ in $G$, against the minimal choice of $K$.
We see, therefore, that $K cap N$ is contained in every maximal subgroup of $K$ and so $K cap N leq Phi(K)$, as wanted.
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
$endgroup$
1
$begingroup$
Thank you very much! Your proof is easy to understand.
$endgroup$
– Zheng Xiao
Nov 1 '18 at 3:06
add a comment |
$begingroup$
Consider the set of supplements to $N$ in $G$. That is, let
$$mathcal{X} = {H leq G : HN = G}.$$
Clearly $mathcal{X}$ is non-empty because $G in mathcal{X}$. Now let $K$ be a minimal element of $mathcal{X}$, in the sense that $K$ does not contain properly any other element from $mathcal{X}$. By definition of $mathcal{X}$ we have $KN = G$. Now let $M$ be a maximal subgroup of $K$. We argue that $K cap N leq M$. Once we have proved that, it will follow that every maximal subgroup of $K$ contains $K cap N$, thus $K cap N leq Phi(K)$.
Suppose for a contradiction that $K cap N$ is not a subgroup of $M$. Since $N$ is normal in $G$, $K cap N$ is normal in $K$. So $K cap N$ permutes with every subgroup of $K$, thus also with $M$. So $(K cap N)M$ is a subgroup of $K$ which contains $M$ properly. Therefore $K = (K cap N)M$. Now notice that
$$G = KN = (K cap N)MN = (K cap N)NM = NM = MN,$$
so $M$ supplements $N$ in $G$, against the minimal choice of $K$.
We see, therefore, that $K cap N$ is contained in every maximal subgroup of $K$ and so $K cap N leq Phi(K)$, as wanted.
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
$endgroup$
1
$begingroup$
Thank you very much! Your proof is easy to understand.
$endgroup$
– Zheng Xiao
Nov 1 '18 at 3:06
add a comment |
$begingroup$
Consider the set of supplements to $N$ in $G$. That is, let
$$mathcal{X} = {H leq G : HN = G}.$$
Clearly $mathcal{X}$ is non-empty because $G in mathcal{X}$. Now let $K$ be a minimal element of $mathcal{X}$, in the sense that $K$ does not contain properly any other element from $mathcal{X}$. By definition of $mathcal{X}$ we have $KN = G$. Now let $M$ be a maximal subgroup of $K$. We argue that $K cap N leq M$. Once we have proved that, it will follow that every maximal subgroup of $K$ contains $K cap N$, thus $K cap N leq Phi(K)$.
Suppose for a contradiction that $K cap N$ is not a subgroup of $M$. Since $N$ is normal in $G$, $K cap N$ is normal in $K$. So $K cap N$ permutes with every subgroup of $K$, thus also with $M$. So $(K cap N)M$ is a subgroup of $K$ which contains $M$ properly. Therefore $K = (K cap N)M$. Now notice that
$$G = KN = (K cap N)MN = (K cap N)NM = NM = MN,$$
so $M$ supplements $N$ in $G$, against the minimal choice of $K$.
We see, therefore, that $K cap N$ is contained in every maximal subgroup of $K$ and so $K cap N leq Phi(K)$, as wanted.
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
$endgroup$
Consider the set of supplements to $N$ in $G$. That is, let
$$mathcal{X} = {H leq G : HN = G}.$$
Clearly $mathcal{X}$ is non-empty because $G in mathcal{X}$. Now let $K$ be a minimal element of $mathcal{X}$, in the sense that $K$ does not contain properly any other element from $mathcal{X}$. By definition of $mathcal{X}$ we have $KN = G$. Now let $M$ be a maximal subgroup of $K$. We argue that $K cap N leq M$. Once we have proved that, it will follow that every maximal subgroup of $K$ contains $K cap N$, thus $K cap N leq Phi(K)$.
Suppose for a contradiction that $K cap N$ is not a subgroup of $M$. Since $N$ is normal in $G$, $K cap N$ is normal in $K$. So $K cap N$ permutes with every subgroup of $K$, thus also with $M$. So $(K cap N)M$ is a subgroup of $K$ which contains $M$ properly. Therefore $K = (K cap N)M$. Now notice that
$$G = KN = (K cap N)MN = (K cap N)NM = NM = MN,$$
so $M$ supplements $N$ in $G$, against the minimal choice of $K$.
We see, therefore, that $K cap N$ is contained in every maximal subgroup of $K$ and so $K cap N leq Phi(K)$, as wanted.
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
edited Oct 31 '18 at 22:53
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
answered Oct 31 '18 at 22:15
the_foxthe_fox
2,90031538
2,90031538
1
$begingroup$
Thank you very much! Your proof is easy to understand.
$endgroup$
– Zheng Xiao
Nov 1 '18 at 3:06
add a comment |
1
$begingroup$
Thank you very much! Your proof is easy to understand.
$endgroup$
– Zheng Xiao
Nov 1 '18 at 3:06
1
1
$begingroup$
Thank you very much! Your proof is easy to understand.
$endgroup$
– Zheng Xiao
Nov 1 '18 at 3:06
$begingroup$
Thank you very much! Your proof is easy to understand.
$endgroup$
– Zheng Xiao
Nov 1 '18 at 3:06
add a comment |
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