Characterization of non solvable groups in terms of normal series.
$begingroup$
I have encountered the following statement which I cannot prove:
A finite group $G$ is not solvable if and only if there is a NORMAL series with
$$ 1lhd H lhd N lhd G $$
such that $N/H$ is a nonabelian simple group or product of isomorphic nonabelian simple groups.
I guess by a normal series is meant the usual, that any term of the series is normal in the whole group $G$.
Any help would be greatly appreciated.
finite-groups normal-subgroups solvable-groups
edited Dec 22 '18 at 5:12
Shaun
9,241113684
asked Jun 5 '17 at 16:09
muser17muser17
192
$endgroup$
add a comment |
$begingroup$
I have encountered the following statement which I cannot prove:
A finite group $G$ is not solvable if and only if there is a NORMAL series with
$$ 1lhd H lhd N lhd G $$
such that $N/H$ is a nonabelian simple group or product of isomorphic nonabelian simple groups.
I guess by a normal series is meant the usual, that any term of the series is normal in the whole group $G$.
Any help would be greatly appreciated.
finite-groups normal-subgroups solvable-groups
edited Dec 22 '18 at 5:12
Shaun
9,241113684
asked Jun 5 '17 at 16:09
muser17muser17
192
$endgroup$
$begingroup$
$H/N$ should be $N/H$. I'm not allowed to do this as an edit as it involves too few characters.
$endgroup$
– David Towers
Jun 6 '17 at 6:54
$begingroup$
Which direction of the proof do you have problems with? In either case, since $G$ is finite, you can use the method of infinite descent (i.e. assume that $G$ is a counterexample of minimal order).
$endgroup$
– jpvee
Jun 8 '17 at 8:46
add a comment |
$begingroup$
I have encountered the following statement which I cannot prove:
A finite group $G$ is not solvable if and only if there is a NORMAL series with
$$ 1lhd H lhd N lhd G $$
such that $N/H$ is a nonabelian simple group or product of isomorphic nonabelian simple groups.
I guess by a normal series is meant the usual, that any term of the series is normal in the whole group $G$.
Any help would be greatly appreciated.
finite-groups normal-subgroups solvable-groups
edited Dec 22 '18 at 5:12
Shaun
9,241113684
asked Jun 5 '17 at 16:09
muser17muser17
192
$endgroup$
I have encountered the following statement which I cannot prove:
A finite group $G$ is not solvable if and only if there is a NORMAL series with
$$ 1lhd H lhd N lhd G $$
such that $N/H$ is a nonabelian simple group or product of isomorphic nonabelian simple groups.
I guess by a normal series is meant the usual, that any term of the series is normal in the whole group $G$.
Any help would be greatly appreciated.
finite-groups normal-subgroups solvable-groups
finite-groups normal-subgroups solvable-groups
edited Dec 22 '18 at 5:12
Shaun
9,241113684
asked Jun 5 '17 at 16:09
muser17muser17
192
edited Dec 22 '18 at 5:12
Shaun
9,241113684
asked Jun 5 '17 at 16:09
muser17muser17
192
edited Dec 22 '18 at 5:12
Shaun
9,241113684
edited Dec 22 '18 at 5:12
Shaun
9,241113684
edited Dec 22 '18 at 5:12
Shaun
9,241113684
9,241113684
asked Jun 5 '17 at 16:09
muser17muser17
192
asked Jun 5 '17 at 16:09
muser17muser17
192
asked Jun 5 '17 at 16:09
muser17muser17
192
192
$begingroup$
$H/N$ should be $N/H$. I'm not allowed to do this as an edit as it involves too few characters.
$endgroup$
– David Towers
Jun 6 '17 at 6:54
$begingroup$
Which direction of the proof do you have problems with? In either case, since $G$ is finite, you can use the method of infinite descent (i.e. assume that $G$ is a counterexample of minimal order).
$endgroup$
– jpvee
Jun 8 '17 at 8:46
add a comment |
$begingroup$
$H/N$ should be $N/H$. I'm not allowed to do this as an edit as it involves too few characters.
$endgroup$
– David Towers
Jun 6 '17 at 6:54
$begingroup$
Which direction of the proof do you have problems with? In either case, since $G$ is finite, you can use the method of infinite descent (i.e. assume that $G$ is a counterexample of minimal order).
$endgroup$
– jpvee
Jun 8 '17 at 8:46
$begingroup$
$H/N$ should be $N/H$. I'm not allowed to do this as an edit as it involves too few characters.
$endgroup$
– David Towers
Jun 6 '17 at 6:54
$begingroup$
$H/N$ should be $N/H$. I'm not allowed to do this as an edit as it involves too few characters.
$endgroup$
– David Towers
Jun 6 '17 at 6:54
$begingroup$
Which direction of the proof do you have problems with? In either case, since $G$ is finite, you can use the method of infinite descent (i.e. assume that $G$ is a counterexample of minimal order).
$endgroup$
– jpvee
Jun 8 '17 at 8:46
$begingroup$
Which direction of the proof do you have problems with? In either case, since $G$ is finite, you can use the method of infinite descent (i.e. assume that $G$ is a counterexample of minimal order).
$endgroup$
– jpvee
Jun 8 '17 at 8:46
add a comment |
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$begingroup$
$H/N$ should be $N/H$. I'm not allowed to do this as an edit as it involves too few characters.
$endgroup$
– David Towers
Jun 6 '17 at 6:54
$begingroup$
Which direction of the proof do you have problems with? In either case, since $G$ is finite, you can use the method of infinite descent (i.e. assume that $G$ is a counterexample of minimal order).
$endgroup$
– jpvee
Jun 8 '17 at 8:46