How to show that $Kleq S_4$ is a normal subgroup?
$begingroup$
Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.
I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.
Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$
group-theory
$endgroup$
add a comment |
$begingroup$
Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.
I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.
Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$
group-theory
$endgroup$
3
$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53
$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57
5
$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56
add a comment |
$begingroup$
Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.
I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.
Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$
group-theory
$endgroup$
Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.
I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.
Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$
group-theory
group-theory
edited Jan 1 at 16:41
Dietrich Burde
80.2k647104
80.2k647104
asked Feb 23 '13 at 15:51
Nancy RutkowskieNancy Rutkowskie
7051718
7051718
3
$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53
$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57
5
$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56
add a comment |
3
$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53
$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57
5
$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56
3
3
$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53
$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53
$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57
$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57
5
5
$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56
$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56
add a comment |
2 Answers
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$begingroup$
This isn't going to work on two fronts:
First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.
Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.
Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.
$endgroup$
add a comment |
$begingroup$
Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This isn't going to work on two fronts:
First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.
Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.
Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.
$endgroup$
add a comment |
$begingroup$
This isn't going to work on two fronts:
First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.
Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.
Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.
$endgroup$
add a comment |
$begingroup$
This isn't going to work on two fronts:
First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.
Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.
Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.
$endgroup$
This isn't going to work on two fronts:
First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.
Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.
Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.
answered Feb 23 '13 at 15:58
JimJim
24.4k23370
24.4k23370
add a comment |
add a comment |
$begingroup$
Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$
$endgroup$
add a comment |
$begingroup$
Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$
$endgroup$
add a comment |
$begingroup$
Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$
$endgroup$
Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$
answered Feb 23 '13 at 16:54
Geoff RobinsonGeoff Robinson
20.6k13043
20.6k13043
add a comment |
add a comment |
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3
$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53
$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57
5
$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56