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Epimorphism onto short exact sequence
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I am studying algebra and now I do not know, if the following is true and why:
Given is the short exact sequence:
$1 to L to L rtimes R to R to 1$.
Additionally there is any group $G$. Moreover there are epimorphisms from $G$ onto $L$ and from $G$ onto $R$. Does this means there exists an epimorphism from $G$ onto $L rtimes R$?
abstract-algebra exact-sequence
asked Jul 18 at 18:24
SHTAM
12
add a comment |Â
up vote
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down vote
favorite
I am studying algebra and now I do not know, if the following is true and why:
Given is the short exact sequence:
$1 to L to L rtimes R to R to 1$.
Additionally there is any group $G$. Moreover there are epimorphisms from $G$ onto $L$ and from $G$ onto $R$. Does this means there exists an epimorphism from $G$ onto $L rtimes R$?
abstract-algebra exact-sequence
asked Jul 18 at 18:24
SHTAM
12
Sounds rather unlikely, what if $L$, $R$ are commutative but the semidirect product isn'?. Then $Ltimes R$ maps onto $L$ and onto $R$ etc.
– Lord Shark the Unknown
Jul 18 at 18:26
1
This isn't true in general, for example the dihedral group of order 8 is a semidirect product of a cyclic group of order 2 and a cyclic group of order 4. So a cyclic group of order 4 maps surjectively onto both but definitely not onto a group of order 8.
– CJD
Jul 18 at 18:45
@CJD You should write that up as an answer.
– Mike Pierce
Jul 18 at 19:17
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I am studying algebra and now I do not know, if the following is true and why:
Given is the short exact sequence:
$1 to L to L rtimes R to R to 1$.
Additionally there is any group $G$. Moreover there are epimorphisms from $G$ onto $L$ and from $G$ onto $R$. Does this means there exists an epimorphism from $G$ onto $L rtimes R$?
abstract-algebra exact-sequence
asked Jul 18 at 18:24
SHTAM
12
I am studying algebra and now I do not know, if the following is true and why:
Given is the short exact sequence:
$1 to L to L rtimes R to R to 1$.
Additionally there is any group $G$. Moreover there are epimorphisms from $G$ onto $L$ and from $G$ onto $R$. Does this means there exists an epimorphism from $G$ onto $L rtimes R$?
abstract-algebra exact-sequence
asked Jul 18 at 18:24
SHTAM
12
asked Jul 18 at 18:24
SHTAM
12
asked Jul 18 at 18:24
SHTAM
12
asked Jul 18 at 18:24
SHTAM
12
12
Sounds rather unlikely, what if $L$, $R$ are commutative but the semidirect product isn'?. Then $Ltimes R$ maps onto $L$ and onto $R$ etc.
– Lord Shark the Unknown
Jul 18 at 18:26
1
This isn't true in general, for example the dihedral group of order 8 is a semidirect product of a cyclic group of order 2 and a cyclic group of order 4. So a cyclic group of order 4 maps surjectively onto both but definitely not onto a group of order 8.
– CJD
Jul 18 at 18:45
@CJD You should write that up as an answer.
– Mike Pierce
Jul 18 at 19:17
add a comment |Â
Sounds rather unlikely, what if $L$, $R$ are commutative but the semidirect product isn'?. Then $Ltimes R$ maps onto $L$ and onto $R$ etc.
– Lord Shark the Unknown
Jul 18 at 18:26
1
This isn't true in general, for example the dihedral group of order 8 is a semidirect product of a cyclic group of order 2 and a cyclic group of order 4. So a cyclic group of order 4 maps surjectively onto both but definitely not onto a group of order 8.
– CJD
Jul 18 at 18:45
@CJD You should write that up as an answer.
– Mike Pierce
Jul 18 at 19:17
Sounds rather unlikely, what if $L$, $R$ are commutative but the semidirect product isn'?. Then $Ltimes R$ maps onto $L$ and onto $R$ etc.
– Lord Shark the Unknown
Jul 18 at 18:26
Sounds rather unlikely, what if $L$, $R$ are commutative but the semidirect product isn'?. Then $Ltimes R$ maps onto $L$ and onto $R$ etc.
– Lord Shark the Unknown
Jul 18 at 18:26
1
1
This isn't true in general, for example the dihedral group of order 8 is a semidirect product of a cyclic group of order 2 and a cyclic group of order 4. So a cyclic group of order 4 maps surjectively onto both but definitely not onto a group of order 8.
– CJD
Jul 18 at 18:45
This isn't true in general, for example the dihedral group of order 8 is a semidirect product of a cyclic group of order 2 and a cyclic group of order 4. So a cyclic group of order 4 maps surjectively onto both but definitely not onto a group of order 8.
– CJD
Jul 18 at 18:45
@CJD You should write that up as an answer.
– Mike Pierce
Jul 18 at 19:17
@CJD You should write that up as an answer.
– Mike Pierce
Jul 18 at 19:17
add a comment |Â
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Sounds rather unlikely, what if $L$, $R$ are commutative but the semidirect product isn'?. Then $Ltimes R$ maps onto $L$ and onto $R$ etc.
– Lord Shark the Unknown
Jul 18 at 18:26
1
This isn't true in general, for example the dihedral group of order 8 is a semidirect product of a cyclic group of order 2 and a cyclic group of order 4. So a cyclic group of order 4 maps surjectively onto both but definitely not onto a group of order 8.
– CJD
Jul 18 at 18:45
@CJD You should write that up as an answer.
– Mike Pierce
Jul 18 at 19:17