A simple question about normal subgroups of finite groups.
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Let $G$ be a finite group and let $H_1 trianglelefteq G$. If $H_2 leq G$, with $|H_1| = |H_2|$, can we state that $H_2 trianglelefteq G$? If it is false in general, is the same statement true supposing furthermore that $H_1$ and $H_2$ are isomorphic?
group-theory finite-groups normal-subgroups group-isomorphism
edited Sep 11 at 17:34
amWhy
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asked Sep 9 at 12:18
joseabp91
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Let $G$ be a finite group and let $H_1 trianglelefteq G$. If $H_2 leq G$, with $|H_1| = |H_2|$, can we state that $H_2 trianglelefteq G$? If it is false in general, is the same statement true supposing furthermore that $H_1$ and $H_2$ are isomorphic?
group-theory finite-groups normal-subgroups group-isomorphism
edited Sep 11 at 17:34
amWhy
190k27221433
asked Sep 9 at 12:18
joseabp91
1,106411
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G$ be a finite group and let $H_1 trianglelefteq G$. If $H_2 leq G$, with $|H_1| = |H_2|$, can we state that $H_2 trianglelefteq G$? If it is false in general, is the same statement true supposing furthermore that $H_1$ and $H_2$ are isomorphic?
group-theory finite-groups normal-subgroups group-isomorphism
edited Sep 11 at 17:34
amWhy
190k27221433
asked Sep 9 at 12:18
joseabp91
1,106411
Let $G$ be a finite group and let $H_1 trianglelefteq G$. If $H_2 leq G$, with $|H_1| = |H_2|$, can we state that $H_2 trianglelefteq G$? If it is false in general, is the same statement true supposing furthermore that $H_1$ and $H_2$ are isomorphic?
group-theory finite-groups normal-subgroups group-isomorphism
group-theory finite-groups normal-subgroups group-isomorphism
edited Sep 11 at 17:34
amWhy
190k27221433
asked Sep 9 at 12:18
joseabp91
1,106411
edited Sep 11 at 17:34
amWhy
190k27221433
asked Sep 9 at 12:18
joseabp91
1,106411
edited Sep 11 at 17:34
amWhy
190k27221433
edited Sep 11 at 17:34
amWhy
190k27221433
edited Sep 11 at 17:34
amWhy
190k27221433
190k27221433
asked Sep 9 at 12:18
joseabp91
1,106411
asked Sep 9 at 12:18
joseabp91
1,106411
asked Sep 9 at 12:18
joseabp91
1,106411
1,106411
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
6
down vote
No.
Pick your favourite example of $Ale B$ with $Anotlhd B$. Then consider $G=Atimes B$ and let $H_1$ be the $A$ that is the first factor and $H_2$ the $A$ contained in the second.
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
I am going to take $tau = (23)$ in the not abelian group of order $6$, $D_3$. Then the order of $tau$ is $2$ and $langle tau rangle$ is a subgroup, but not normal in $D_3$. Then can I take $G = langle tau rangle times D_3$? This group must have order $12$. Then do $H_1 = langle tau rangle times Id cong mathbbZ_2$ and $H_2 = Id times langle tau rangle cong mathbbZ_2$ work as counterexample?
– joseabp91
Sep 9 at 12:50
add a comment |Â
up vote
2
down vote
For the first argument there is a well-known which results: If $$|G/H_1|=2$$ where $|G|<infty$ so, $|G/H_2|=2$ so $H_2$ would be a normal subgroup. So this, as you stated, may not be true in general. Moreover, we could have the first statement true when $p$ be the smallest prime number which divides $|G|$ and $|G/H|=p$.
answered Sep 9 at 12:29
mrs
58.5k750143
You do not need to assume $|G|<infty$ for this result. The proof goes: If $H$ has index two in $G$ then there are two left cosets $H, aH$, two right cosets $H, Ha$, and left and right cosets are the same sets. Hence, $aH=Ha$, and so $Hlhd G$.
– user1729
Sep 12 at 9:33
@user1729: Thanks for consideration! Yes you are right. Maybe I wanted to walk in OP's track who had assumed the group is finite.
– mrs
Sep 12 at 17:59
add a comment |Â
up vote
1
down vote
No. Take the group of all $2times2$ matrix over $mathbbF_3$ (or any other finite field whose characteristic is not $2$). Now, consider $H_1=pmoperatornameId$ and$$H_2=leftoperatornameId,beginbmatrix1&0\0&-1endbmatrixright.$$
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
add a comment |Â
up vote
1
down vote
Several subgroups of the symmetric group $S_4$ are isomorphic to the Klein Four-Group, but only one of them is a normal subgroup.
answered Sep 9 at 12:36
bof
47k349114
And if I suppose furthermore that $|G| leq 15$?
– joseabp91
Sep 9 at 12:40
@joseabp91 Sure. $S_2times S_3$ is a group of order $12$ which has seven subgroups of order two, and just one of them is normal.
– bof
Sep 10 at 8:13
add a comment |Â
up vote
0
down vote
It is true also if $H_1$ is a Sylow subgroup of $G$ (this is the Second Sylow theorem).
answered Sep 9 at 12:33
Bernard
112k636105
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
No.
Pick your favourite example of $Ale B$ with $Anotlhd B$. Then consider $G=Atimes B$ and let $H_1$ be the $A$ that is the first factor and $H_2$ the $A$ contained in the second.
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
I am going to take $tau = (23)$ in the not abelian group of order $6$, $D_3$. Then the order of $tau$ is $2$ and $langle tau rangle$ is a subgroup, but not normal in $D_3$. Then can I take $G = langle tau rangle times D_3$? This group must have order $12$. Then do $H_1 = langle tau rangle times Id cong mathbbZ_2$ and $H_2 = Id times langle tau rangle cong mathbbZ_2$ work as counterexample?
– joseabp91
Sep 9 at 12:50
add a comment |Â
up vote
6
down vote
No.
Pick your favourite example of $Ale B$ with $Anotlhd B$. Then consider $G=Atimes B$ and let $H_1$ be the $A$ that is the first factor and $H_2$ the $A$ contained in the second.
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
I am going to take $tau = (23)$ in the not abelian group of order $6$, $D_3$. Then the order of $tau$ is $2$ and $langle tau rangle$ is a subgroup, but not normal in $D_3$. Then can I take $G = langle tau rangle times D_3$? This group must have order $12$. Then do $H_1 = langle tau rangle times Id cong mathbbZ_2$ and $H_2 = Id times langle tau rangle cong mathbbZ_2$ work as counterexample?
– joseabp91
Sep 9 at 12:50
add a comment |Â
up vote
6
down vote
up vote
6
down vote
No.
Pick your favourite example of $Ale B$ with $Anotlhd B$. Then consider $G=Atimes B$ and let $H_1$ be the $A$ that is the first factor and $H_2$ the $A$ contained in the second.
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
No.
Pick your favourite example of $Ale B$ with $Anotlhd B$. Then consider $G=Atimes B$ and let $H_1$ be the $A$ that is the first factor and $H_2$ the $A$ contained in the second.
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
answered Sep 9 at 12:22
Hagen von Eitzen
267k21260484
267k21260484
I am going to take $tau = (23)$ in the not abelian group of order $6$, $D_3$. Then the order of $tau$ is $2$ and $langle tau rangle$ is a subgroup, but not normal in $D_3$. Then can I take $G = langle tau rangle times D_3$? This group must have order $12$. Then do $H_1 = langle tau rangle times Id cong mathbbZ_2$ and $H_2 = Id times langle tau rangle cong mathbbZ_2$ work as counterexample?
– joseabp91
Sep 9 at 12:50
add a comment |Â
I am going to take $tau = (23)$ in the not abelian group of order $6$, $D_3$. Then the order of $tau$ is $2$ and $langle tau rangle$ is a subgroup, but not normal in $D_3$. Then can I take $G = langle tau rangle times D_3$? This group must have order $12$. Then do $H_1 = langle tau rangle times Id cong mathbbZ_2$ and $H_2 = Id times langle tau rangle cong mathbbZ_2$ work as counterexample?
– joseabp91
Sep 9 at 12:50
I am going to take $tau = (23)$ in the not abelian group of order $6$, $D_3$. Then the order of $tau$ is $2$ and $langle tau rangle$ is a subgroup, but not normal in $D_3$. Then can I take $G = langle tau rangle times D_3$? This group must have order $12$. Then do $H_1 = langle tau rangle times Id cong mathbbZ_2$ and $H_2 = Id times langle tau rangle cong mathbbZ_2$ work as counterexample?
– joseabp91
Sep 9 at 12:50
I am going to take $tau = (23)$ in the not abelian group of order $6$, $D_3$. Then the order of $tau$ is $2$ and $langle tau rangle$ is a subgroup, but not normal in $D_3$. Then can I take $G = langle tau rangle times D_3$? This group must have order $12$. Then do $H_1 = langle tau rangle times Id cong mathbbZ_2$ and $H_2 = Id times langle tau rangle cong mathbbZ_2$ work as counterexample?
– joseabp91
Sep 9 at 12:50
add a comment |Â
up vote
2
down vote
For the first argument there is a well-known which results: If $$|G/H_1|=2$$ where $|G|<infty$ so, $|G/H_2|=2$ so $H_2$ would be a normal subgroup. So this, as you stated, may not be true in general. Moreover, we could have the first statement true when $p$ be the smallest prime number which divides $|G|$ and $|G/H|=p$.
answered Sep 9 at 12:29
mrs
58.5k750143
You do not need to assume $|G|<infty$ for this result. The proof goes: If $H$ has index two in $G$ then there are two left cosets $H, aH$, two right cosets $H, Ha$, and left and right cosets are the same sets. Hence, $aH=Ha$, and so $Hlhd G$.
– user1729
Sep 12 at 9:33
@user1729: Thanks for consideration! Yes you are right. Maybe I wanted to walk in OP's track who had assumed the group is finite.
– mrs
Sep 12 at 17:59
add a comment |Â
up vote
2
down vote
For the first argument there is a well-known which results: If $$|G/H_1|=2$$ where $|G|<infty$ so, $|G/H_2|=2$ so $H_2$ would be a normal subgroup. So this, as you stated, may not be true in general. Moreover, we could have the first statement true when $p$ be the smallest prime number which divides $|G|$ and $|G/H|=p$.
answered Sep 9 at 12:29
mrs
58.5k750143
You do not need to assume $|G|<infty$ for this result. The proof goes: If $H$ has index two in $G$ then there are two left cosets $H, aH$, two right cosets $H, Ha$, and left and right cosets are the same sets. Hence, $aH=Ha$, and so $Hlhd G$.
– user1729
Sep 12 at 9:33
@user1729: Thanks for consideration! Yes you are right. Maybe I wanted to walk in OP's track who had assumed the group is finite.
– mrs
Sep 12 at 17:59
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For the first argument there is a well-known which results: If $$|G/H_1|=2$$ where $|G|<infty$ so, $|G/H_2|=2$ so $H_2$ would be a normal subgroup. So this, as you stated, may not be true in general. Moreover, we could have the first statement true when $p$ be the smallest prime number which divides $|G|$ and $|G/H|=p$.
answered Sep 9 at 12:29
mrs
58.5k750143
For the first argument there is a well-known which results: If $$|G/H_1|=2$$ where $|G|<infty$ so, $|G/H_2|=2$ so $H_2$ would be a normal subgroup. So this, as you stated, may not be true in general. Moreover, we could have the first statement true when $p$ be the smallest prime number which divides $|G|$ and $|G/H|=p$.
answered Sep 9 at 12:29
mrs
58.5k750143
edited Sep 9 at 12:35
answered Sep 9 at 12:29
mrs
58.5k750143
answered Sep 9 at 12:29
mrs
58.5k750143
answered Sep 9 at 12:29
mrs
58.5k750143
58.5k750143
You do not need to assume $|G|<infty$ for this result. The proof goes: If $H$ has index two in $G$ then there are two left cosets $H, aH$, two right cosets $H, Ha$, and left and right cosets are the same sets. Hence, $aH=Ha$, and so $Hlhd G$.
– user1729
Sep 12 at 9:33
@user1729: Thanks for consideration! Yes you are right. Maybe I wanted to walk in OP's track who had assumed the group is finite.
– mrs
Sep 12 at 17:59
add a comment |Â
You do not need to assume $|G|<infty$ for this result. The proof goes: If $H$ has index two in $G$ then there are two left cosets $H, aH$, two right cosets $H, Ha$, and left and right cosets are the same sets. Hence, $aH=Ha$, and so $Hlhd G$.
– user1729
Sep 12 at 9:33
@user1729: Thanks for consideration! Yes you are right. Maybe I wanted to walk in OP's track who had assumed the group is finite.
– mrs
Sep 12 at 17:59
You do not need to assume $|G|<infty$ for this result. The proof goes: If $H$ has index two in $G$ then there are two left cosets $H, aH$, two right cosets $H, Ha$, and left and right cosets are the same sets. Hence, $aH=Ha$, and so $Hlhd G$.
– user1729
Sep 12 at 9:33
You do not need to assume $|G|<infty$ for this result. The proof goes: If $H$ has index two in $G$ then there are two left cosets $H, aH$, two right cosets $H, Ha$, and left and right cosets are the same sets. Hence, $aH=Ha$, and so $Hlhd G$.
– user1729
Sep 12 at 9:33
@user1729: Thanks for consideration! Yes you are right. Maybe I wanted to walk in OP's track who had assumed the group is finite.
– mrs
Sep 12 at 17:59
@user1729: Thanks for consideration! Yes you are right. Maybe I wanted to walk in OP's track who had assumed the group is finite.
– mrs
Sep 12 at 17:59
add a comment |Â
up vote
1
down vote
No. Take the group of all $2times2$ matrix over $mathbbF_3$ (or any other finite field whose characteristic is not $2$). Now, consider $H_1=pmoperatornameId$ and$$H_2=leftoperatornameId,beginbmatrix1&0\0&-1endbmatrixright.$$
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
add a comment |Â
up vote
1
down vote
No. Take the group of all $2times2$ matrix over $mathbbF_3$ (or any other finite field whose characteristic is not $2$). Now, consider $H_1=pmoperatornameId$ and$$H_2=leftoperatornameId,beginbmatrix1&0\0&-1endbmatrixright.$$
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No. Take the group of all $2times2$ matrix over $mathbbF_3$ (or any other finite field whose characteristic is not $2$). Now, consider $H_1=pmoperatornameId$ and$$H_2=leftoperatornameId,beginbmatrix1&0\0&-1endbmatrixright.$$
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
No. Take the group of all $2times2$ matrix over $mathbbF_3$ (or any other finite field whose characteristic is not $2$). Now, consider $H_1=pmoperatornameId$ and$$H_2=leftoperatornameId,beginbmatrix1&0\0&-1endbmatrixright.$$
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
answered Sep 9 at 12:31
José Carlos Santos
123k17101186
123k17101186
add a comment |Â
add a comment |Â
up vote
1
down vote
Several subgroups of the symmetric group $S_4$ are isomorphic to the Klein Four-Group, but only one of them is a normal subgroup.
answered Sep 9 at 12:36
bof
47k349114
And if I suppose furthermore that $|G| leq 15$?
– joseabp91
Sep 9 at 12:40
@joseabp91 Sure. $S_2times S_3$ is a group of order $12$ which has seven subgroups of order two, and just one of them is normal.
– bof
Sep 10 at 8:13
add a comment |Â
up vote
1
down vote
Several subgroups of the symmetric group $S_4$ are isomorphic to the Klein Four-Group, but only one of them is a normal subgroup.
answered Sep 9 at 12:36
bof
47k349114
And if I suppose furthermore that $|G| leq 15$?
– joseabp91
Sep 9 at 12:40
@joseabp91 Sure. $S_2times S_3$ is a group of order $12$ which has seven subgroups of order two, and just one of them is normal.
– bof
Sep 10 at 8:13
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Several subgroups of the symmetric group $S_4$ are isomorphic to the Klein Four-Group, but only one of them is a normal subgroup.
answered Sep 9 at 12:36
bof
47k349114
Several subgroups of the symmetric group $S_4$ are isomorphic to the Klein Four-Group, but only one of them is a normal subgroup.
answered Sep 9 at 12:36
bof
47k349114
answered Sep 9 at 12:36
bof
47k349114
answered Sep 9 at 12:36
bof
47k349114
answered Sep 9 at 12:36
bof
47k349114
47k349114
And if I suppose furthermore that $|G| leq 15$?
– joseabp91
Sep 9 at 12:40
@joseabp91 Sure. $S_2times S_3$ is a group of order $12$ which has seven subgroups of order two, and just one of them is normal.
– bof
Sep 10 at 8:13
add a comment |Â
And if I suppose furthermore that $|G| leq 15$?
– joseabp91
Sep 9 at 12:40
@joseabp91 Sure. $S_2times S_3$ is a group of order $12$ which has seven subgroups of order two, and just one of them is normal.
– bof
Sep 10 at 8:13
And if I suppose furthermore that $|G| leq 15$?
– joseabp91
Sep 9 at 12:40
And if I suppose furthermore that $|G| leq 15$?
– joseabp91
Sep 9 at 12:40
@joseabp91 Sure. $S_2times S_3$ is a group of order $12$ which has seven subgroups of order two, and just one of them is normal.
– bof
Sep 10 at 8:13
@joseabp91 Sure. $S_2times S_3$ is a group of order $12$ which has seven subgroups of order two, and just one of them is normal.
– bof
Sep 10 at 8:13
add a comment |Â
up vote
0
down vote
It is true also if $H_1$ is a Sylow subgroup of $G$ (this is the Second Sylow theorem).
answered Sep 9 at 12:33
Bernard
112k636105
add a comment |Â
up vote
0
down vote
It is true also if $H_1$ is a Sylow subgroup of $G$ (this is the Second Sylow theorem).
answered Sep 9 at 12:33
Bernard
112k636105
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is true also if $H_1$ is a Sylow subgroup of $G$ (this is the Second Sylow theorem).
answered Sep 9 at 12:33
Bernard
112k636105
It is true also if $H_1$ is a Sylow subgroup of $G$ (this is the Second Sylow theorem).
answered Sep 9 at 12:33
Bernard
112k636105
answered Sep 9 at 12:33
Bernard
112k636105
answered Sep 9 at 12:33
Bernard
112k636105
answered Sep 9 at 12:33
Bernard
112k636105
112k636105
add a comment |Â
add a comment |Â
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