Subgroup generated by a subgroup and a conjugate of itCalculating presentations for the normal subgroup of a semidirect productExistence of simultaneously normal finite index subgroupsNormal intermediate subgroup and normal coreNormal subgroup lattice of the group $U_6n$Is there a big solvable subgroup in every finite group?Free generators for the fat commutator subgroupProper subgroups which have the same minimal subgroups(revision)Existence of a cyclic non-normal subgroup in a $p$-groupSubgroup embedding properties paranormality and polynormality
Subgroup generated by a subgroup and a conjugate of it
Calculating presentations for the normal subgroup of a semidirect productExistence of simultaneously normal finite index subgroupsNormal intermediate subgroup and normal coreNormal subgroup lattice of the group $U_6n$Is there a big solvable subgroup in every finite group?Free generators for the fat commutator subgroupProper subgroups which have the same minimal subgroups(revision)Existence of a cyclic non-normal subgroup in a $p$-groupSubgroup embedding properties paranormality and polynormality
$begingroup$
Let $Hleq G$ be groups, and $ain G$ so that $langle H,arangle=G$. Does it follows that $langle Hcup aHa^-1rangle$ is a normal subgroup of $G$?
My hope is that this is true, and my guess is that it is not.
It might be easy.
gr.group-theory finite-groups normal-subgroups
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
asked 9 hours ago
Bertalan BodorBertalan Bodor
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New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Let $Hleq G$ be groups, and $ain G$ so that $langle H,arangle=G$. Does it follows that $langle Hcup aHa^-1rangle$ is a normal subgroup of $G$?
My hope is that this is true, and my guess is that it is not.
It might be easy.
gr.group-theory finite-groups normal-subgroups
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
asked 9 hours ago
Bertalan BodorBertalan Bodor
161 bronze badge
New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
An obvious counterexample is when $G=Hwrlangle arangle$ with $a$ of order $ge 3$ (this gives a counterexample of order $24$ with $H$ of order $2$, $a$ of order $3$).
$endgroup$
– YCor
6 hours ago
1
$begingroup$
Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
$begingroup$
Let $Hleq G$ be groups, and $ain G$ so that $langle H,arangle=G$. Does it follows that $langle Hcup aHa^-1rangle$ is a normal subgroup of $G$?
My hope is that this is true, and my guess is that it is not.
It might be easy.
gr.group-theory finite-groups normal-subgroups
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
asked 9 hours ago
Bertalan BodorBertalan Bodor
161 bronze badge
New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $Hleq G$ be groups, and $ain G$ so that $langle H,arangle=G$. Does it follows that $langle Hcup aHa^-1rangle$ is a normal subgroup of $G$?
My hope is that this is true, and my guess is that it is not.
It might be easy.
gr.group-theory finite-groups normal-subgroups
gr.group-theory finite-groups normal-subgroups
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
asked 9 hours ago
Bertalan BodorBertalan Bodor
161 bronze badge
New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
asked 9 hours ago
Bertalan BodorBertalan Bodor
161 bronze badge
New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
edited 6 hours ago
YCor
31k4 gold badges95 silver badges147 bronze badges
31k4 gold badges95 silver badges147 bronze badges
asked 9 hours ago
Bertalan BodorBertalan Bodor
161 bronze badge
New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Bertalan BodorBertalan Bodor
161 bronze badge
asked 9 hours ago
Bertalan BodorBertalan Bodor
161 bronze badge
161 bronze badge
New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bertalan Bodor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
An obvious counterexample is when $G=Hwrlangle arangle$ with $a$ of order $ge 3$ (this gives a counterexample of order $24$ with $H$ of order $2$, $a$ of order $3$).
$endgroup$
– YCor
6 hours ago
1
$begingroup$
Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
2
$begingroup$
An obvious counterexample is when $G=Hwrlangle arangle$ with $a$ of order $ge 3$ (this gives a counterexample of order $24$ with $H$ of order $2$, $a$ of order $3$).
$endgroup$
– YCor
6 hours ago
1
$begingroup$
Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this.
$endgroup$
– Bertalan Bodor
4 hours ago
2
2
$begingroup$
An obvious counterexample is when $G=Hwrlangle arangle$ with $a$ of order $ge 3$ (this gives a counterexample of order $24$ with $H$ of order $2$, $a$ of order $3$).
$endgroup$
– YCor
6 hours ago
$begingroup$
An obvious counterexample is when $G=Hwrlangle arangle$ with $a$ of order $ge 3$ (this gives a counterexample of order $24$ with $H$ of order $2$, $a$ of order $3$).
$endgroup$
– YCor
6 hours ago
1
1
$begingroup$
Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this.
$endgroup$
– Bertalan Bodor
4 hours ago
$begingroup$
Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Too long for a comment. The answer is no.
Just take the free product $G=H*langle a rangle$, where $anotin H$ is of infinite order. By definition, $H$ and $a$ generates $G$.
Then $langle Hcup aHa^-1rangle $ is not normal.
For example, letting $hin H$, $g:=a^2ha^-2notin langle Hcup aHa^-1rangle$. Indeed by contradiction, write $g=h_1...h_n$, where $h_kin H$ or $h_kin aHa^-1$. Then, the first letter needs to be an $a$ so that $h_1in aHa^-1$ but the second letter of $h_1$ is in $H$, whereas the second letter of $g$ is again $a$.
It seems to me that the problem actually comes from the powers of $a$. For example, assume that $a^2in H$, then, the answer is yes. Indeed you then have that $G/H$ is of order 2, so that $H$ is normal in $G$ and so $langle Hcup aHa^-1rangle =H$ is indeed normal. So if you have a particular group in mind, maybe you should first look carefully as the relations between powers of $a$ and $H$.
Also quick comment: try to find more suited titles.
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
$endgroup$
$begingroup$
Thanks, very nice.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
$begingroup$
As other have said, the answer is no in general. However if $a$ has finite order $n$, then it is true that $langle H cup aHa^-1 cup ldots cup a^n-1Ha^-(n-1) rangle$ is a normal subgroup of $G$. This is because the group in question is normalized by $H$ (since it contains $H$ ) and is normalized by $a$, since the given generating set is stable under conjugation by $a$ ( and hence the group it generates is stable under conjugation by $a$). Hence the given group is normalized by $langle H, a rangle = G$, so is a normal subgroup of $G$.
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
Work in $GL_2(mathbbQ)$. Let $H$ be the subgroup generated by
$$
left(beginarray 01 & 1\ 0 & 1endarrayright).
$$
Let $G=langle H,arangle$ where
$$
a=left(beginarray 02 & 0\ 0 & 1endarrayright).
$$
Then $aHa^-1$ is a proper subgroup of $H$. In particular, $H$ is not normal in $G$, and $H=langle Hcup aHa^-1rangle$.
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
$endgroup$
add a comment |
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3 Answers
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active
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votes
3 Answers
3
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votes
$begingroup$
Too long for a comment. The answer is no.
Just take the free product $G=H*langle a rangle$, where $anotin H$ is of infinite order. By definition, $H$ and $a$ generates $G$.
Then $langle Hcup aHa^-1rangle $ is not normal.
For example, letting $hin H$, $g:=a^2ha^-2notin langle Hcup aHa^-1rangle$. Indeed by contradiction, write $g=h_1...h_n$, where $h_kin H$ or $h_kin aHa^-1$. Then, the first letter needs to be an $a$ so that $h_1in aHa^-1$ but the second letter of $h_1$ is in $H$, whereas the second letter of $g$ is again $a$.
It seems to me that the problem actually comes from the powers of $a$. For example, assume that $a^2in H$, then, the answer is yes. Indeed you then have that $G/H$ is of order 2, so that $H$ is normal in $G$ and so $langle Hcup aHa^-1rangle =H$ is indeed normal. So if you have a particular group in mind, maybe you should first look carefully as the relations between powers of $a$ and $H$.
Also quick comment: try to find more suited titles.
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
$endgroup$
$begingroup$
Thanks, very nice.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
$begingroup$
Too long for a comment. The answer is no.
Just take the free product $G=H*langle a rangle$, where $anotin H$ is of infinite order. By definition, $H$ and $a$ generates $G$.
Then $langle Hcup aHa^-1rangle $ is not normal.
For example, letting $hin H$, $g:=a^2ha^-2notin langle Hcup aHa^-1rangle$. Indeed by contradiction, write $g=h_1...h_n$, where $h_kin H$ or $h_kin aHa^-1$. Then, the first letter needs to be an $a$ so that $h_1in aHa^-1$ but the second letter of $h_1$ is in $H$, whereas the second letter of $g$ is again $a$.
It seems to me that the problem actually comes from the powers of $a$. For example, assume that $a^2in H$, then, the answer is yes. Indeed you then have that $G/H$ is of order 2, so that $H$ is normal in $G$ and so $langle Hcup aHa^-1rangle =H$ is indeed normal. So if you have a particular group in mind, maybe you should first look carefully as the relations between powers of $a$ and $H$.
Also quick comment: try to find more suited titles.
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
$endgroup$
$begingroup$
Thanks, very nice.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
$begingroup$
Too long for a comment. The answer is no.
Just take the free product $G=H*langle a rangle$, where $anotin H$ is of infinite order. By definition, $H$ and $a$ generates $G$.
Then $langle Hcup aHa^-1rangle $ is not normal.
For example, letting $hin H$, $g:=a^2ha^-2notin langle Hcup aHa^-1rangle$. Indeed by contradiction, write $g=h_1...h_n$, where $h_kin H$ or $h_kin aHa^-1$. Then, the first letter needs to be an $a$ so that $h_1in aHa^-1$ but the second letter of $h_1$ is in $H$, whereas the second letter of $g$ is again $a$.
It seems to me that the problem actually comes from the powers of $a$. For example, assume that $a^2in H$, then, the answer is yes. Indeed you then have that $G/H$ is of order 2, so that $H$ is normal in $G$ and so $langle Hcup aHa^-1rangle =H$ is indeed normal. So if you have a particular group in mind, maybe you should first look carefully as the relations between powers of $a$ and $H$.
Also quick comment: try to find more suited titles.
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
$endgroup$
Too long for a comment. The answer is no.
Just take the free product $G=H*langle a rangle$, where $anotin H$ is of infinite order. By definition, $H$ and $a$ generates $G$.
Then $langle Hcup aHa^-1rangle $ is not normal.
For example, letting $hin H$, $g:=a^2ha^-2notin langle Hcup aHa^-1rangle$. Indeed by contradiction, write $g=h_1...h_n$, where $h_kin H$ or $h_kin aHa^-1$. Then, the first letter needs to be an $a$ so that $h_1in aHa^-1$ but the second letter of $h_1$ is in $H$, whereas the second letter of $g$ is again $a$.
It seems to me that the problem actually comes from the powers of $a$. For example, assume that $a^2in H$, then, the answer is yes. Indeed you then have that $G/H$ is of order 2, so that $H$ is normal in $G$ and so $langle Hcup aHa^-1rangle =H$ is indeed normal. So if you have a particular group in mind, maybe you should first look carefully as the relations between powers of $a$ and $H$.
Also quick comment: try to find more suited titles.
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
answered 8 hours ago
M. DusM. Dus
6421 gold badge4 silver badges21 bronze badges
6421 gold badge4 silver badges21 bronze badges
$begingroup$
Thanks, very nice.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
$begingroup$
Thanks, very nice.
$endgroup$
– Bertalan Bodor
4 hours ago
$begingroup$
Thanks, very nice.
$endgroup$
– Bertalan Bodor
4 hours ago
$begingroup$
Thanks, very nice.
$endgroup$
– Bertalan Bodor
4 hours ago
add a comment |
$begingroup$
As other have said, the answer is no in general. However if $a$ has finite order $n$, then it is true that $langle H cup aHa^-1 cup ldots cup a^n-1Ha^-(n-1) rangle$ is a normal subgroup of $G$. This is because the group in question is normalized by $H$ (since it contains $H$ ) and is normalized by $a$, since the given generating set is stable under conjugation by $a$ ( and hence the group it generates is stable under conjugation by $a$). Hence the given group is normalized by $langle H, a rangle = G$, so is a normal subgroup of $G$.
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
As other have said, the answer is no in general. However if $a$ has finite order $n$, then it is true that $langle H cup aHa^-1 cup ldots cup a^n-1Ha^-(n-1) rangle$ is a normal subgroup of $G$. This is because the group in question is normalized by $H$ (since it contains $H$ ) and is normalized by $a$, since the given generating set is stable under conjugation by $a$ ( and hence the group it generates is stable under conjugation by $a$). Hence the given group is normalized by $langle H, a rangle = G$, so is a normal subgroup of $G$.
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
As other have said, the answer is no in general. However if $a$ has finite order $n$, then it is true that $langle H cup aHa^-1 cup ldots cup a^n-1Ha^-(n-1) rangle$ is a normal subgroup of $G$. This is because the group in question is normalized by $H$ (since it contains $H$ ) and is normalized by $a$, since the given generating set is stable under conjugation by $a$ ( and hence the group it generates is stable under conjugation by $a$). Hence the given group is normalized by $langle H, a rangle = G$, so is a normal subgroup of $G$.
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
$endgroup$
As other have said, the answer is no in general. However if $a$ has finite order $n$, then it is true that $langle H cup aHa^-1 cup ldots cup a^n-1Ha^-(n-1) rangle$ is a normal subgroup of $G$. This is because the group in question is normalized by $H$ (since it contains $H$ ) and is normalized by $a$, since the given generating set is stable under conjugation by $a$ ( and hence the group it generates is stable under conjugation by $a$). Hence the given group is normalized by $langle H, a rangle = G$, so is a normal subgroup of $G$.
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
edited 7 hours ago
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
answered 7 hours ago
Geoff RobinsonGeoff Robinson
30.8k2 gold badges85 silver badges117 bronze badges
30.8k2 gold badges85 silver badges117 bronze badges
add a comment |
add a comment |
$begingroup$
Work in $GL_2(mathbbQ)$. Let $H$ be the subgroup generated by
$$
left(beginarray 01 & 1\ 0 & 1endarrayright).
$$
Let $G=langle H,arangle$ where
$$
a=left(beginarray 02 & 0\ 0 & 1endarrayright).
$$
Then $aHa^-1$ is a proper subgroup of $H$. In particular, $H$ is not normal in $G$, and $H=langle Hcup aHa^-1rangle$.
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Work in $GL_2(mathbbQ)$. Let $H$ be the subgroup generated by
$$
left(beginarray 01 & 1\ 0 & 1endarrayright).
$$
Let $G=langle H,arangle$ where
$$
a=left(beginarray 02 & 0\ 0 & 1endarrayright).
$$
Then $aHa^-1$ is a proper subgroup of $H$. In particular, $H$ is not normal in $G$, and $H=langle Hcup aHa^-1rangle$.
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Work in $GL_2(mathbbQ)$. Let $H$ be the subgroup generated by
$$
left(beginarray 01 & 1\ 0 & 1endarrayright).
$$
Let $G=langle H,arangle$ where
$$
a=left(beginarray 02 & 0\ 0 & 1endarrayright).
$$
Then $aHa^-1$ is a proper subgroup of $H$. In particular, $H$ is not normal in $G$, and $H=langle Hcup aHa^-1rangle$.
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
$endgroup$
Work in $GL_2(mathbbQ)$. Let $H$ be the subgroup generated by
$$
left(beginarray 01 & 1\ 0 & 1endarrayright).
$$
Let $G=langle H,arangle$ where
$$
a=left(beginarray 02 & 0\ 0 & 1endarrayright).
$$
Then $aHa^-1$ is a proper subgroup of $H$. In particular, $H$ is not normal in $G$, and $H=langle Hcup aHa^-1rangle$.
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
edited 8 hours ago
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
answered 8 hours ago
Gabe ConantGabe Conant
1,20710 silver badges17 bronze badges
1,20710 silver badges17 bronze badges
add a comment |
add a comment |
Bertalan Bodor is a new contributor. Be nice, and check out our Code of Conduct.
Bertalan Bodor is a new contributor. Be nice, and check out our Code of Conduct.
Bertalan Bodor is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
An obvious counterexample is when $G=Hwrlangle arangle$ with $a$ of order $ge 3$ (this gives a counterexample of order $24$ with $H$ of order $2$, $a$ of order $3$).
$endgroup$
– YCor
6 hours ago
1
$begingroup$
Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this.
$endgroup$
– Bertalan Bodor
4 hours ago