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Let $A,B$ be solvable subgroup of a group $G$, suppose $B subset N_G(A)$. Prove $AB$ is solvable [on hold]
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Show: Let $A,B$ be solvable subgroup of a group $G$. Suppose the $B subset N_G(A)$. Prove that $AB$ is solvable.
abstract-algebra group-theory
edited yesterday
darij grinberg
9,89532961
asked yesterday
mathnoob
63311
put on hold as off-topic by Derek Holt, jgon, user10354138, Cesareo, Shailesh yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, jgon, user10354138, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
1
down vote
favorite
Need help,
Show: Let $A,B$ be solvable subgroup of a group $G$. Suppose the $B subset N_G(A)$. Prove that $AB$ is solvable.
abstract-algebra group-theory
edited yesterday
darij grinberg
9,89532961
asked yesterday
mathnoob
63311
put on hold as off-topic by Derek Holt, jgon, user10354138, Cesareo, Shailesh yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, jgon, user10354138, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
What do you know? What have you tried?
– verret
yesterday
Someone gave me a hint to show that AB is a subgroup of G, and based on the fact that any subgroup of a solvable group is solvable, AB must be solvable. AB is a subgroup because aba'b'=aa''bb'. Hence closed under product. (ab)^{-1}=b^{-1}a^{-1}=a'b^{-1} hence closed under inversion. So, I just have to read why any subgroup of a solvable group is solvable.
– mathnoob
yesterday
Well if there’s $G=G_0 supseteq G_1 supseteq ... supseteq G_n={e} $-solvable series for G, then just consider such series for $H subseteq G : H_i=G_i cap H$ and try to prove that this series will be solvable for $H$
– Anton Zagrivin
yesterday
Hint: you’ll need a second isomorphism theorem
– Anton Zagrivin
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Need help,
Show: Let $A,B$ be solvable subgroup of a group $G$. Suppose the $B subset N_G(A)$. Prove that $AB$ is solvable.
abstract-algebra group-theory
edited yesterday
darij grinberg
9,89532961
asked yesterday
mathnoob
63311
Need help,
Show: Let $A,B$ be solvable subgroup of a group $G$. Suppose the $B subset N_G(A)$. Prove that $AB$ is solvable.
abstract-algebra group-theory
abstract-algebra group-theory
edited yesterday
darij grinberg
9,89532961
asked yesterday
mathnoob
63311
edited yesterday
darij grinberg
9,89532961
asked yesterday
mathnoob
63311
edited yesterday
darij grinberg
9,89532961
edited yesterday
darij grinberg
9,89532961
edited yesterday
darij grinberg
9,89532961
9,89532961
asked yesterday
mathnoob
63311
asked yesterday
mathnoob
63311
asked yesterday
mathnoob
63311
63311
put on hold as off-topic by Derek Holt, jgon, user10354138, Cesareo, Shailesh yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, jgon, user10354138, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Derek Holt, jgon, user10354138, Cesareo, Shailesh yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, jgon, user10354138, Cesareo, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
What do you know? What have you tried?
– verret
yesterday
Someone gave me a hint to show that AB is a subgroup of G, and based on the fact that any subgroup of a solvable group is solvable, AB must be solvable. AB is a subgroup because aba'b'=aa''bb'. Hence closed under product. (ab)^{-1}=b^{-1}a^{-1}=a'b^{-1} hence closed under inversion. So, I just have to read why any subgroup of a solvable group is solvable.
– mathnoob
yesterday
Well if there’s $G=G_0 supseteq G_1 supseteq ... supseteq G_n={e} $-solvable series for G, then just consider such series for $H subseteq G : H_i=G_i cap H$ and try to prove that this series will be solvable for $H$
– Anton Zagrivin
yesterday
Hint: you’ll need a second isomorphism theorem
– Anton Zagrivin
yesterday
add a comment |
2
What do you know? What have you tried?
– verret
yesterday
Someone gave me a hint to show that AB is a subgroup of G, and based on the fact that any subgroup of a solvable group is solvable, AB must be solvable. AB is a subgroup because aba'b'=aa''bb'. Hence closed under product. (ab)^{-1}=b^{-1}a^{-1}=a'b^{-1} hence closed under inversion. So, I just have to read why any subgroup of a solvable group is solvable.
– mathnoob
yesterday
Well if there’s $G=G_0 supseteq G_1 supseteq ... supseteq G_n={e} $-solvable series for G, then just consider such series for $H subseteq G : H_i=G_i cap H$ and try to prove that this series will be solvable for $H$
– Anton Zagrivin
yesterday
Hint: you’ll need a second isomorphism theorem
– Anton Zagrivin
yesterday
2
2
What do you know? What have you tried?
– verret
yesterday
What do you know? What have you tried?
– verret
yesterday
Someone gave me a hint to show that AB is a subgroup of G, and based on the fact that any subgroup of a solvable group is solvable, AB must be solvable. AB is a subgroup because aba'b'=aa''bb'. Hence closed under product. (ab)^{-1}=b^{-1}a^{-1}=a'b^{-1} hence closed under inversion. So, I just have to read why any subgroup of a solvable group is solvable.
– mathnoob
yesterday
Someone gave me a hint to show that AB is a subgroup of G, and based on the fact that any subgroup of a solvable group is solvable, AB must be solvable. AB is a subgroup because aba'b'=aa''bb'. Hence closed under product. (ab)^{-1}=b^{-1}a^{-1}=a'b^{-1} hence closed under inversion. So, I just have to read why any subgroup of a solvable group is solvable.
– mathnoob
yesterday
Well if there’s $G=G_0 supseteq G_1 supseteq ... supseteq G_n={e} $-solvable series for G, then just consider such series for $H subseteq G : H_i=G_i cap H$ and try to prove that this series will be solvable for $H$
– Anton Zagrivin
yesterday
Well if there’s $G=G_0 supseteq G_1 supseteq ... supseteq G_n={e} $-solvable series for G, then just consider such series for $H subseteq G : H_i=G_i cap H$ and try to prove that this series will be solvable for $H$
– Anton Zagrivin
yesterday
Hint: you’ll need a second isomorphism theorem
– Anton Zagrivin
yesterday
Hint: you’ll need a second isomorphism theorem
– Anton Zagrivin
yesterday
add a comment |
1 Answer
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up vote
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accepted
Because $B subseteq N_G(A)$, it is easy to show that $AB$ is actually a subgroup. Since $B$ normalizes $A$ it follows that $A unlhd AB$. Now $AB/A cong B/(A cap B)$ by the 2nd isomorphism theorem. Since $A$ is solvable and $B$, whence any of $B$'s quotients is solvable, it follows that $AB$ is solvable. (In general: $G$ is solvable iff $G/N$ and $N$ are solvable ($N unlhd G$)).
answered yesterday
Nicky Hekster
27.9k53254
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Because $B subseteq N_G(A)$, it is easy to show that $AB$ is actually a subgroup. Since $B$ normalizes $A$ it follows that $A unlhd AB$. Now $AB/A cong B/(A cap B)$ by the 2nd isomorphism theorem. Since $A$ is solvable and $B$, whence any of $B$'s quotients is solvable, it follows that $AB$ is solvable. (In general: $G$ is solvable iff $G/N$ and $N$ are solvable ($N unlhd G$)).
answered yesterday
Nicky Hekster
27.9k53254
add a comment |
up vote
2
down vote
accepted
Because $B subseteq N_G(A)$, it is easy to show that $AB$ is actually a subgroup. Since $B$ normalizes $A$ it follows that $A unlhd AB$. Now $AB/A cong B/(A cap B)$ by the 2nd isomorphism theorem. Since $A$ is solvable and $B$, whence any of $B$'s quotients is solvable, it follows that $AB$ is solvable. (In general: $G$ is solvable iff $G/N$ and $N$ are solvable ($N unlhd G$)).
answered yesterday
Nicky Hekster
27.9k53254
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Because $B subseteq N_G(A)$, it is easy to show that $AB$ is actually a subgroup. Since $B$ normalizes $A$ it follows that $A unlhd AB$. Now $AB/A cong B/(A cap B)$ by the 2nd isomorphism theorem. Since $A$ is solvable and $B$, whence any of $B$'s quotients is solvable, it follows that $AB$ is solvable. (In general: $G$ is solvable iff $G/N$ and $N$ are solvable ($N unlhd G$)).
answered yesterday
Nicky Hekster
27.9k53254
Because $B subseteq N_G(A)$, it is easy to show that $AB$ is actually a subgroup. Since $B$ normalizes $A$ it follows that $A unlhd AB$. Now $AB/A cong B/(A cap B)$ by the 2nd isomorphism theorem. Since $A$ is solvable and $B$, whence any of $B$'s quotients is solvable, it follows that $AB$ is solvable. (In general: $G$ is solvable iff $G/N$ and $N$ are solvable ($N unlhd G$)).
answered yesterday
Nicky Hekster
27.9k53254
edited yesterday
answered yesterday
Nicky Hekster
27.9k53254
answered yesterday
Nicky Hekster
27.9k53254
answered yesterday
Nicky Hekster
27.9k53254
27.9k53254
add a comment |
add a comment |
2
What do you know? What have you tried?
– verret
yesterday
Someone gave me a hint to show that AB is a subgroup of G, and based on the fact that any subgroup of a solvable group is solvable, AB must be solvable. AB is a subgroup because aba'b'=aa''bb'. Hence closed under product. (ab)^{-1}=b^{-1}a^{-1}=a'b^{-1} hence closed under inversion. So, I just have to read why any subgroup of a solvable group is solvable.
– mathnoob
yesterday
Well if there’s $G=G_0 supseteq G_1 supseteq ... supseteq G_n={e} $-solvable series for G, then just consider such series for $H subseteq G : H_i=G_i cap H$ and try to prove that this series will be solvable for $H$
– Anton Zagrivin
yesterday
Hint: you’ll need a second isomorphism theorem
– Anton Zagrivin
yesterday