Mixing
Prove that if $X,Ytriangleleft G$ and $G=XY$ that $Xcap Y={e}$ then $Gcong Xtimes Y$. [duplicate]
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This question already has an answer here:
Prove $KNcong Ktimes N$ if $Kcap N={e}$
1 answer
I had a midterm exam today about group theory and I were asked to prove the following theorem:
Let $G$ be a group. Prove that if $X,Ytriangleleft G$ and $G=XY$ that $Xcap Y={e}$ then $Gcong Xtimes Y$.
I really have no idea where to start proving this theorem. Also I tried to find a previous thread about this question but without any luck.
How to prove this theorem?
group-theory
asked Jan 8 at 0:25
BadukBaduk
171
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marked as duplicate by Trevor Gunn, A. Pongrácz, Saad, KReiser, Leucippus Jan 8 at 3:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Prove $KNcong Ktimes N$ if $Kcap N={e}$
1 answer
I had a midterm exam today about group theory and I were asked to prove the following theorem:
Let $G$ be a group. Prove that if $X,Ytriangleleft G$ and $G=XY$ that $Xcap Y={e}$ then $Gcong Xtimes Y$.
I really have no idea where to start proving this theorem. Also I tried to find a previous thread about this question but without any luck.
How to prove this theorem?
group-theory
asked Jan 8 at 0:25
BadukBaduk
171
$endgroup$
marked as duplicate by Trevor Gunn, A. Pongrácz, Saad, KReiser, Leucippus Jan 8 at 3:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
It must be a duplicate. However, a hint: show that each element $g$ of $G$ can be uniquely written as $g=xy$ with $xin X, , yin Y$, and show that this mapping $gmapsto (x, y)$ is a group homomorphism. The key is that $[x, y] in Xcap Y$.
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– Berci
Jan 8 at 0:41
add a comment |
$begingroup$
This question already has an answer here:
Prove $KNcong Ktimes N$ if $Kcap N={e}$
1 answer
I had a midterm exam today about group theory and I were asked to prove the following theorem:
Let $G$ be a group. Prove that if $X,Ytriangleleft G$ and $G=XY$ that $Xcap Y={e}$ then $Gcong Xtimes Y$.
I really have no idea where to start proving this theorem. Also I tried to find a previous thread about this question but without any luck.
How to prove this theorem?
group-theory
asked Jan 8 at 0:25
BadukBaduk
171
$endgroup$
This question already has an answer here:
Prove $KNcong Ktimes N$ if $Kcap N={e}$
1 answer
I had a midterm exam today about group theory and I were asked to prove the following theorem:
Let $G$ be a group. Prove that if $X,Ytriangleleft G$ and $G=XY$ that $Xcap Y={e}$ then $Gcong Xtimes Y$.
I really have no idea where to start proving this theorem. Also I tried to find a previous thread about this question but without any luck.
How to prove this theorem?
This question already has an answer here:
Prove $KNcong Ktimes N$ if $Kcap N={e}$
1 answer
group-theory
group-theory
asked Jan 8 at 0:25
BadukBaduk
171
asked Jan 8 at 0:25
BadukBaduk
171
asked Jan 8 at 0:25
BadukBaduk
171
asked Jan 8 at 0:25
BadukBaduk
171
asked Jan 8 at 0:25
BadukBaduk
171
171
marked as duplicate by Trevor Gunn, A. Pongrácz, Saad, KReiser, Leucippus Jan 8 at 3:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Trevor Gunn, A. Pongrácz, Saad, KReiser, Leucippus Jan 8 at 3:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
It must be a duplicate. However, a hint: show that each element $g$ of $G$ can be uniquely written as $g=xy$ with $xin X, , yin Y$, and show that this mapping $gmapsto (x, y)$ is a group homomorphism. The key is that $[x, y] in Xcap Y$.
$endgroup$
– Berci
Jan 8 at 0:41
add a comment |
$begingroup$
It must be a duplicate. However, a hint: show that each element $g$ of $G$ can be uniquely written as $g=xy$ with $xin X, , yin Y$, and show that this mapping $gmapsto (x, y)$ is a group homomorphism. The key is that $[x, y] in Xcap Y$.
$endgroup$
– Berci
Jan 8 at 0:41
$begingroup$
It must be a duplicate. However, a hint: show that each element $g$ of $G$ can be uniquely written as $g=xy$ with $xin X, , yin Y$, and show that this mapping $gmapsto (x, y)$ is a group homomorphism. The key is that $[x, y] in Xcap Y$.
$endgroup$
– Berci
Jan 8 at 0:41
$begingroup$
It must be a duplicate. However, a hint: show that each element $g$ of $G$ can be uniquely written as $g=xy$ with $xin X, , yin Y$, and show that this mapping $gmapsto (x, y)$ is a group homomorphism. The key is that $[x, y] in Xcap Y$.
$endgroup$
– Berci
Jan 8 at 0:41
add a comment |
1 Answer
1
active
oldest
votes
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Hint: prove first that $(x,y) in X times Y longmapsto xy$ is a bijection. Then, note that $xyx’y’=(xx’)((x’^{-1}yx’)y’)=(x(yx’y^{-1}))(yy’)$, thus $(x’^{-1}yx’)y’=yy’$ (Because we have a bijection and the subgroups are normal) and thus the bijection is a homomorphism.
answered Jan 8 at 0:44
MindlackMindlack
3,45717
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can you explain a bit more? I don't see it for some reason.
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– vesii
Jan 8 at 0:46
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What is your issue?
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– Mindlack
Jan 8 at 0:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: prove first that $(x,y) in X times Y longmapsto xy$ is a bijection. Then, note that $xyx’y’=(xx’)((x’^{-1}yx’)y’)=(x(yx’y^{-1}))(yy’)$, thus $(x’^{-1}yx’)y’=yy’$ (Because we have a bijection and the subgroups are normal) and thus the bijection is a homomorphism.
answered Jan 8 at 0:44
MindlackMindlack
3,45717
$endgroup$
$begingroup$
can you explain a bit more? I don't see it for some reason.
$endgroup$
– vesii
Jan 8 at 0:46
$begingroup$
What is your issue?
$endgroup$
– Mindlack
Jan 8 at 0:48
add a comment |
$begingroup$
Hint: prove first that $(x,y) in X times Y longmapsto xy$ is a bijection. Then, note that $xyx’y’=(xx’)((x’^{-1}yx’)y’)=(x(yx’y^{-1}))(yy’)$, thus $(x’^{-1}yx’)y’=yy’$ (Because we have a bijection and the subgroups are normal) and thus the bijection is a homomorphism.
answered Jan 8 at 0:44
MindlackMindlack
3,45717
$endgroup$
$begingroup$
can you explain a bit more? I don't see it for some reason.
$endgroup$
– vesii
Jan 8 at 0:46
$begingroup$
What is your issue?
$endgroup$
– Mindlack
Jan 8 at 0:48
add a comment |
$begingroup$
Hint: prove first that $(x,y) in X times Y longmapsto xy$ is a bijection. Then, note that $xyx’y’=(xx’)((x’^{-1}yx’)y’)=(x(yx’y^{-1}))(yy’)$, thus $(x’^{-1}yx’)y’=yy’$ (Because we have a bijection and the subgroups are normal) and thus the bijection is a homomorphism.
answered Jan 8 at 0:44
MindlackMindlack
3,45717
$endgroup$
Hint: prove first that $(x,y) in X times Y longmapsto xy$ is a bijection. Then, note that $xyx’y’=(xx’)((x’^{-1}yx’)y’)=(x(yx’y^{-1}))(yy’)$, thus $(x’^{-1}yx’)y’=yy’$ (Because we have a bijection and the subgroups are normal) and thus the bijection is a homomorphism.
answered Jan 8 at 0:44
MindlackMindlack
3,45717
answered Jan 8 at 0:44
MindlackMindlack
3,45717
answered Jan 8 at 0:44
MindlackMindlack
3,45717
answered Jan 8 at 0:44
MindlackMindlack
3,45717
3,45717
$begingroup$
can you explain a bit more? I don't see it for some reason.
$endgroup$
– vesii
Jan 8 at 0:46
$begingroup$
What is your issue?
$endgroup$
– Mindlack
Jan 8 at 0:48
add a comment |
$begingroup$
can you explain a bit more? I don't see it for some reason.
$endgroup$
– vesii
Jan 8 at 0:46
$begingroup$
What is your issue?
$endgroup$
– Mindlack
Jan 8 at 0:48
$begingroup$
can you explain a bit more? I don't see it for some reason.
$endgroup$
– vesii
Jan 8 at 0:46
$begingroup$
can you explain a bit more? I don't see it for some reason.
$endgroup$
– vesii
Jan 8 at 0:46
$begingroup$
What is your issue?
$endgroup$
– Mindlack
Jan 8 at 0:48
$begingroup$
What is your issue?
$endgroup$
– Mindlack
Jan 8 at 0:48
add a comment |
$begingroup$
It must be a duplicate. However, a hint: show that each element $g$ of $G$ can be uniquely written as $g=xy$ with $xin X, , yin Y$, and show that this mapping $gmapsto (x, y)$ is a group homomorphism. The key is that $[x, y] in Xcap Y$.
$endgroup$
– Berci
Jan 8 at 0:41