Any Subgroup containing commutator subgroup is normal.
Clash Royale CLAN TAG#URR8PPP up vote
6
down vote
favorite
I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as
If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.
I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$
It is for elements in $G'$ to show condition for normality.
But how to do for elements not in $G'$ but in $H$ that in $H/G'$?
abstract-algebra group-theory normal-subgroups
edited yesterday
Marcus Müller
1827
asked yesterday
SRJ
984216
add a comment |Â
up vote
6
down vote
favorite
I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as
If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.
I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$
It is for elements in $G'$ to show condition for normality.
But how to do for elements not in $G'$ but in $H$ that in $H/G'$?
abstract-algebra group-theory normal-subgroups
edited yesterday
Marcus Müller
1827
asked yesterday
SRJ
984216
1
See this question.
– Dietrich Burde
yesterday
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as
If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.
I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$
It is for elements in $G'$ to show condition for normality.
But how to do for elements not in $G'$ but in $H$ that in $H/G'$?
abstract-algebra group-theory normal-subgroups
edited yesterday
Marcus Müller
1827
asked yesterday
SRJ
984216
I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as
If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.
I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$
It is for elements in $G'$ to show condition for normality.
But how to do for elements not in $G'$ but in $H$ that in $H/G'$?
abstract-algebra group-theory normal-subgroups
edited yesterday
Marcus Müller
1827
asked yesterday
SRJ
984216
edited yesterday
Marcus Müller
1827
edited yesterday
Marcus Müller
1827
edited yesterday
Marcus Müller
1827
1827
asked yesterday
SRJ
984216
asked yesterday
SRJ
984216
asked yesterday
SRJ
984216
984216
1
See this question.
– Dietrich Burde
yesterday
add a comment |Â
1
See this question.
– Dietrich Burde
yesterday
1
1
See this question.
– Dietrich Burde
yesterday
See this question.
– Dietrich Burde
yesterday
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
11
down vote
accepted
If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.
answered yesterday
José Carlos Santos
111k1695171
add a comment |Â
up vote
9
down vote
$G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
isomorphism theorem for groups, $H$ is normal in $G$.
answered yesterday
Lord Shark the Unknown
83.8k949111
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.
answered yesterday
José Carlos Santos
111k1695171
add a comment |Â
up vote
11
down vote
accepted
If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.
answered yesterday
José Carlos Santos
111k1695171
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.
answered yesterday
José Carlos Santos
111k1695171
If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.
answered yesterday
José Carlos Santos
111k1695171
edited yesterday
answered yesterday
José Carlos Santos
111k1695171
answered yesterday
José Carlos Santos
111k1695171
answered yesterday
José Carlos Santos
111k1695171
111k1695171
add a comment |Â
add a comment |Â
up vote
9
down vote
$G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
isomorphism theorem for groups, $H$ is normal in $G$.
answered yesterday
Lord Shark the Unknown
83.8k949111
add a comment |Â
up vote
9
down vote
$G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
isomorphism theorem for groups, $H$ is normal in $G$.
answered yesterday
Lord Shark the Unknown
83.8k949111
add a comment |Â
up vote
9
down vote
up vote
9
down vote
$G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
isomorphism theorem for groups, $H$ is normal in $G$.
answered yesterday
Lord Shark the Unknown
83.8k949111
$G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
isomorphism theorem for groups, $H$ is normal in $G$.
answered yesterday
Lord Shark the Unknown
83.8k949111
answered yesterday
Lord Shark the Unknown
83.8k949111
answered yesterday
Lord Shark the Unknown
83.8k949111
answered yesterday
Lord Shark the Unknown
83.8k949111
83.8k949111
add a comment |Â
add a comment |Â
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1
See this question.
– Dietrich Burde
yesterday