Mixing
About normal minimal subgroups not in the Frattini
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
In Neukirch--Schmidt--Wingberg, "Cohomology of Number Fields", Second edition, page 624, Exercise 2, it is stated the following fact.
$textbfClaim$: If $N$ is a normal subgroup, minimal among normal subgroups, of a group $G$ not contained in the Frattini's subgroup of $G$, then $G$ is a semi-direct product over $N$, i.e. there exists $H$ subgroup of $G$ with $H cap N=id$ and $HN=G$.
Now, if $N$ is a simple non-abelian normal subgroup of $G$, the above Claim implies that $G$ is a semi-direct product over $N$. Indeed the Frattini of $G$ is nilpotent, and since $N$ is non-abelian, it cannot be contained in a nilpotent group, otherwise $N$ itself would be nilpotent, and nilpotent and simple implies having prime order and hence abelian. Moreover $N$ is clearly minimal, since $N$ itself does not have non-trivial proper normal subgroup. But then I see a clear contradiction with Lemma 4.3 of this paper of Lucchini, Menegazzo and Morigi https://projecteuclid.org/download/pdf_1/euclid.ijm/1258488162.
Could somebody explain (in case I did not perform a mistake in reasoning) which of the two publications has a mistake?
nt.number-theory gr.group-theory finite-groups abstract-algebra
edited 37 mins ago
Alexey Ustinov
6,47245777
asked 1 hour ago
Tom1909
211
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
favorite
In Neukirch--Schmidt--Wingberg, "Cohomology of Number Fields", Second edition, page 624, Exercise 2, it is stated the following fact.
$textbfClaim$: If $N$ is a normal subgroup, minimal among normal subgroups, of a group $G$ not contained in the Frattini's subgroup of $G$, then $G$ is a semi-direct product over $N$, i.e. there exists $H$ subgroup of $G$ with $H cap N=id$ and $HN=G$.
Now, if $N$ is a simple non-abelian normal subgroup of $G$, the above Claim implies that $G$ is a semi-direct product over $N$. Indeed the Frattini of $G$ is nilpotent, and since $N$ is non-abelian, it cannot be contained in a nilpotent group, otherwise $N$ itself would be nilpotent, and nilpotent and simple implies having prime order and hence abelian. Moreover $N$ is clearly minimal, since $N$ itself does not have non-trivial proper normal subgroup. But then I see a clear contradiction with Lemma 4.3 of this paper of Lucchini, Menegazzo and Morigi https://projecteuclid.org/download/pdf_1/euclid.ijm/1258488162.
Could somebody explain (in case I did not perform a mistake in reasoning) which of the two publications has a mistake?
nt.number-theory gr.group-theory finite-groups abstract-algebra
edited 37 mins ago
Alexey Ustinov
6,47245777
asked 1 hour ago
Tom1909
211
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In Neukirch--Schmidt--Wingberg, "Cohomology of Number Fields", Second edition, page 624, Exercise 2, it is stated the following fact.
$textbfClaim$: If $N$ is a normal subgroup, minimal among normal subgroups, of a group $G$ not contained in the Frattini's subgroup of $G$, then $G$ is a semi-direct product over $N$, i.e. there exists $H$ subgroup of $G$ with $H cap N=id$ and $HN=G$.
Now, if $N$ is a simple non-abelian normal subgroup of $G$, the above Claim implies that $G$ is a semi-direct product over $N$. Indeed the Frattini of $G$ is nilpotent, and since $N$ is non-abelian, it cannot be contained in a nilpotent group, otherwise $N$ itself would be nilpotent, and nilpotent and simple implies having prime order and hence abelian. Moreover $N$ is clearly minimal, since $N$ itself does not have non-trivial proper normal subgroup. But then I see a clear contradiction with Lemma 4.3 of this paper of Lucchini, Menegazzo and Morigi https://projecteuclid.org/download/pdf_1/euclid.ijm/1258488162.
Could somebody explain (in case I did not perform a mistake in reasoning) which of the two publications has a mistake?
nt.number-theory gr.group-theory finite-groups abstract-algebra
edited 37 mins ago
Alexey Ustinov
6,47245777
asked 1 hour ago
Tom1909
211
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In Neukirch--Schmidt--Wingberg, "Cohomology of Number Fields", Second edition, page 624, Exercise 2, it is stated the following fact.
$textbfClaim$: If $N$ is a normal subgroup, minimal among normal subgroups, of a group $G$ not contained in the Frattini's subgroup of $G$, then $G$ is a semi-direct product over $N$, i.e. there exists $H$ subgroup of $G$ with $H cap N=id$ and $HN=G$.
Now, if $N$ is a simple non-abelian normal subgroup of $G$, the above Claim implies that $G$ is a semi-direct product over $N$. Indeed the Frattini of $G$ is nilpotent, and since $N$ is non-abelian, it cannot be contained in a nilpotent group, otherwise $N$ itself would be nilpotent, and nilpotent and simple implies having prime order and hence abelian. Moreover $N$ is clearly minimal, since $N$ itself does not have non-trivial proper normal subgroup. But then I see a clear contradiction with Lemma 4.3 of this paper of Lucchini, Menegazzo and Morigi https://projecteuclid.org/download/pdf_1/euclid.ijm/1258488162.
Could somebody explain (in case I did not perform a mistake in reasoning) which of the two publications has a mistake?
nt.number-theory gr.group-theory finite-groups abstract-algebra
nt.number-theory gr.group-theory finite-groups abstract-algebra
edited 37 mins ago
Alexey Ustinov
6,47245777
asked 1 hour ago
Tom1909
211
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 37 mins ago
Alexey Ustinov
6,47245777
asked 1 hour ago
Tom1909
211
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 37 mins ago
Alexey Ustinov
6,47245777
edited 37 mins ago
Alexey Ustinov
6,47245777
edited 37 mins ago
Alexey Ustinov
6,47245777
6,47245777
asked 1 hour ago
Tom1909
211
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Tom1909
211
asked 1 hour ago
Tom1909
211
211
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tom1909 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
I think it is the claim from exercise 2 of Neukirch et al which is incorrect. I presume that all groups are meant to be finite in this question. The claim from exercise 2 is OK when $N$ is solvable (for then $N$ is an elementary Abelian $p$-group for some prime $p$ and there is a maximal subgroup $H$ of $G$ which does not contain $N$. Then $H cap N$ is normal in $N$ ( as $N$ is Abelian), and also normal in $H$, so $H cap N$ is normal in $langle H,N rangle= G.$ By minimality of $N$ among (non-trivial) normal subgroups, we have either $H cap N = N$ or $H cap N =1.$ But we can't have $H cap N = N$ as $H$ does not contain $N$).
On the other hand, there are finite non-Abelian simple groups which are not complemented in their automorphism groups, and the example of Lucchini et al is one of these). Any such example contradicts the claim of Neukirch et al ( if indeed they do not have the restriction that $N$ is assumed solvable).
answered 51 mins ago
Geoff Robinson
28.4k277107
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
I think it is the claim from exercise 2 of Neukirch et al which is incorrect. I presume that all groups are meant to be finite in this question. The claim from exercise 2 is OK when $N$ is solvable (for then $N$ is an elementary Abelian $p$-group for some prime $p$ and there is a maximal subgroup $H$ of $G$ which does not contain $N$. Then $H cap N$ is normal in $N$ ( as $N$ is Abelian), and also normal in $H$, so $H cap N$ is normal in $langle H,N rangle= G.$ By minimality of $N$ among (non-trivial) normal subgroups, we have either $H cap N = N$ or $H cap N =1.$ But we can't have $H cap N = N$ as $H$ does not contain $N$).
On the other hand, there are finite non-Abelian simple groups which are not complemented in their automorphism groups, and the example of Lucchini et al is one of these). Any such example contradicts the claim of Neukirch et al ( if indeed they do not have the restriction that $N$ is assumed solvable).
answered 51 mins ago
Geoff Robinson
28.4k277107
add a comment |Â
up vote
4
down vote
I think it is the claim from exercise 2 of Neukirch et al which is incorrect. I presume that all groups are meant to be finite in this question. The claim from exercise 2 is OK when $N$ is solvable (for then $N$ is an elementary Abelian $p$-group for some prime $p$ and there is a maximal subgroup $H$ of $G$ which does not contain $N$. Then $H cap N$ is normal in $N$ ( as $N$ is Abelian), and also normal in $H$, so $H cap N$ is normal in $langle H,N rangle= G.$ By minimality of $N$ among (non-trivial) normal subgroups, we have either $H cap N = N$ or $H cap N =1.$ But we can't have $H cap N = N$ as $H$ does not contain $N$).
On the other hand, there are finite non-Abelian simple groups which are not complemented in their automorphism groups, and the example of Lucchini et al is one of these). Any such example contradicts the claim of Neukirch et al ( if indeed they do not have the restriction that $N$ is assumed solvable).
answered 51 mins ago
Geoff Robinson
28.4k277107
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I think it is the claim from exercise 2 of Neukirch et al which is incorrect. I presume that all groups are meant to be finite in this question. The claim from exercise 2 is OK when $N$ is solvable (for then $N$ is an elementary Abelian $p$-group for some prime $p$ and there is a maximal subgroup $H$ of $G$ which does not contain $N$. Then $H cap N$ is normal in $N$ ( as $N$ is Abelian), and also normal in $H$, so $H cap N$ is normal in $langle H,N rangle= G.$ By minimality of $N$ among (non-trivial) normal subgroups, we have either $H cap N = N$ or $H cap N =1.$ But we can't have $H cap N = N$ as $H$ does not contain $N$).
On the other hand, there are finite non-Abelian simple groups which are not complemented in their automorphism groups, and the example of Lucchini et al is one of these). Any such example contradicts the claim of Neukirch et al ( if indeed they do not have the restriction that $N$ is assumed solvable).
answered 51 mins ago
Geoff Robinson
28.4k277107
I think it is the claim from exercise 2 of Neukirch et al which is incorrect. I presume that all groups are meant to be finite in this question. The claim from exercise 2 is OK when $N$ is solvable (for then $N$ is an elementary Abelian $p$-group for some prime $p$ and there is a maximal subgroup $H$ of $G$ which does not contain $N$. Then $H cap N$ is normal in $N$ ( as $N$ is Abelian), and also normal in $H$, so $H cap N$ is normal in $langle H,N rangle= G.$ By minimality of $N$ among (non-trivial) normal subgroups, we have either $H cap N = N$ or $H cap N =1.$ But we can't have $H cap N = N$ as $H$ does not contain $N$).
On the other hand, there are finite non-Abelian simple groups which are not complemented in their automorphism groups, and the example of Lucchini et al is one of these). Any such example contradicts the claim of Neukirch et al ( if indeed they do not have the restriction that $N$ is assumed solvable).
answered 51 mins ago
Geoff Robinson
28.4k277107
edited 12 mins ago
answered 51 mins ago
Geoff Robinson
28.4k277107
answered 51 mins ago
Geoff Robinson
28.4k277107
answered 51 mins ago
Geoff Robinson
28.4k277107
28.4k277107
add a comment |Â
add a comment |Â
Tom1909 is a new contributor. Be nice, and check out our Code of Conduct.
Tom1909 is a new contributor. Be nice, and check out our Code of Conduct.
Tom1909 is a new contributor. Be nice, and check out our Code of Conduct.
Tom1909 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f312286%2fabout-normal-minimal-subgroups-not-in-the-frattini%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
