Proving ${gY:gin X}={Yg:gin X}$ [duplicate]
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This question already has an answer here:
How to show that $Kleq S_4$ is a normal subgroup?
2 answers
I was trying the study a new subject in groups (from a book I have). I came acorss with the following question:
Consider:
$X=S_4$
$Y={e,(12)(34),(13)(24),(14)(23)}$
How can I prove that:
$${gY:gin X}={Yg:gin X}$$
group-theory symmetric-groups
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
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marked as duplicate by Dietrich Burde group-theory
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Jan 1 at 16:38
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:
How to show that $Kleq S_4$ is a normal subgroup?
2 answers
I was trying the study a new subject in groups (from a book I have). I came acorss with the following question:
Consider:
$X=S_4$
$Y={e,(12)(34),(13)(24),(14)(23)}$
How can I prove that:
$${gY:gin X}={Yg:gin X}$$
group-theory symmetric-groups
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
$endgroup$
marked as duplicate by Dietrich Burde group-theory
Users with the group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.
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Jan 1 at 16:38
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.
2
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And what is $H$?
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– Fakemistake
Jan 1 at 16:32
1
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Hey guys, sorry I copied the question not right. I have edited.
$endgroup$
– TTaJTa4
Jan 1 at 16:36
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$Y$ is a normal subgroup of $S_4$, which means $gY=Yg$ for all $gin S_4$. See the answer by Geoff Robinson.
$endgroup$
– Dietrich Burde
Jan 1 at 16:38
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@TTaJTa4 thank you for fixing the question!
$endgroup$
– layman
Jan 1 at 16:39
add a comment |
$begingroup$
This question already has an answer here:
How to show that $Kleq S_4$ is a normal subgroup?
2 answers
I was trying the study a new subject in groups (from a book I have). I came acorss with the following question:
Consider:
$X=S_4$
$Y={e,(12)(34),(13)(24),(14)(23)}$
How can I prove that:
$${gY:gin X}={Yg:gin X}$$
group-theory symmetric-groups
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
$endgroup$
This question already has an answer here:
How to show that $Kleq S_4$ is a normal subgroup?
2 answers
I was trying the study a new subject in groups (from a book I have). I came acorss with the following question:
Consider:
$X=S_4$
$Y={e,(12)(34),(13)(24),(14)(23)}$
How can I prove that:
$${gY:gin X}={Yg:gin X}$$
This question already has an answer here:
How to show that $Kleq S_4$ is a normal subgroup?
2 answers
group-theory symmetric-groups
group-theory symmetric-groups
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
edited Jan 1 at 16:35
TTaJTa4
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
asked Jan 1 at 16:30
TTaJTa4TTaJTa4
1666
1666
marked as duplicate by Dietrich Burde group-theory
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Jan 1 at 16:38
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 Dietrich Burde group-theory
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Jan 1 at 16:38
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.
2
$begingroup$
And what is $H$?
$endgroup$
– Fakemistake
Jan 1 at 16:32
1
$begingroup$
Hey guys, sorry I copied the question not right. I have edited.
$endgroup$
– TTaJTa4
Jan 1 at 16:36
$begingroup$
$Y$ is a normal subgroup of $S_4$, which means $gY=Yg$ for all $gin S_4$. See the answer by Geoff Robinson.
$endgroup$
– Dietrich Burde
Jan 1 at 16:38
$begingroup$
@TTaJTa4 thank you for fixing the question!
$endgroup$
– layman
Jan 1 at 16:39
add a comment |
2
$begingroup$
And what is $H$?
$endgroup$
– Fakemistake
Jan 1 at 16:32
1
$begingroup$
Hey guys, sorry I copied the question not right. I have edited.
$endgroup$
– TTaJTa4
Jan 1 at 16:36
$begingroup$
$Y$ is a normal subgroup of $S_4$, which means $gY=Yg$ for all $gin S_4$. See the answer by Geoff Robinson.
$endgroup$
– Dietrich Burde
Jan 1 at 16:38
$begingroup$
@TTaJTa4 thank you for fixing the question!
$endgroup$
– layman
Jan 1 at 16:39
2
2
$begingroup$
And what is $H$?
$endgroup$
– Fakemistake
Jan 1 at 16:32
$begingroup$
And what is $H$?
$endgroup$
– Fakemistake
Jan 1 at 16:32
1
1
$begingroup$
Hey guys, sorry I copied the question not right. I have edited.
$endgroup$
– TTaJTa4
Jan 1 at 16:36
$begingroup$
Hey guys, sorry I copied the question not right. I have edited.
$endgroup$
– TTaJTa4
Jan 1 at 16:36
$begingroup$
$Y$ is a normal subgroup of $S_4$, which means $gY=Yg$ for all $gin S_4$. See the answer by Geoff Robinson.
$endgroup$
– Dietrich Burde
Jan 1 at 16:38
$begingroup$
$Y$ is a normal subgroup of $S_4$, which means $gY=Yg$ for all $gin S_4$. See the answer by Geoff Robinson.
$endgroup$
– Dietrich Burde
Jan 1 at 16:38
$begingroup$
@TTaJTa4 thank you for fixing the question!
$endgroup$
– layman
Jan 1 at 16:39
$begingroup$
@TTaJTa4 thank you for fixing the question!
$endgroup$
– layman
Jan 1 at 16:39
add a comment |
1 Answer
1
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oldest
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$begingroup$
To prove $GY subset YG$, you need to prove that for all $g_1 in G$ and $y_1 in Y$, there exists $y_2 in G$ and $y_2 in Y$ such that $g_1 y_1 = y_2 g_2$, which is equivalent to saying such that $g_1 y_1 g_2^{-1} in Y$. Proving the other inclusion is similar.
We can actually prove a stronger property : for all $g in G$ and $y in Y$, we have $gyg^{-1} in Y$.
Indeed, if $y=e$ we have $gyg^{-1}=ein Y$. Otherwise, $y$ is of the form $y=(a , b) (c , d)$ (with $a, b, c, d$ numbers from $1$ to $4$ (in some order). We compute that
$$g y g^{-1} = (g(a) , g(b)) (g(c) , g(d))$$
And we check that this is still in $Y$ (note that $Y$ contains all permutations of the form $(a , b) (c , d)$, because for example $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To prove $GY subset YG$, you need to prove that for all $g_1 in G$ and $y_1 in Y$, there exists $y_2 in G$ and $y_2 in Y$ such that $g_1 y_1 = y_2 g_2$, which is equivalent to saying such that $g_1 y_1 g_2^{-1} in Y$. Proving the other inclusion is similar.
We can actually prove a stronger property : for all $g in G$ and $y in Y$, we have $gyg^{-1} in Y$.
Indeed, if $y=e$ we have $gyg^{-1}=ein Y$. Otherwise, $y$ is of the form $y=(a , b) (c , d)$ (with $a, b, c, d$ numbers from $1$ to $4$ (in some order). We compute that
$$g y g^{-1} = (g(a) , g(b)) (g(c) , g(d))$$
And we check that this is still in $Y$ (note that $Y$ contains all permutations of the form $(a , b) (c , d)$, because for example $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
$endgroup$
add a comment |
$begingroup$
To prove $GY subset YG$, you need to prove that for all $g_1 in G$ and $y_1 in Y$, there exists $y_2 in G$ and $y_2 in Y$ such that $g_1 y_1 = y_2 g_2$, which is equivalent to saying such that $g_1 y_1 g_2^{-1} in Y$. Proving the other inclusion is similar.
We can actually prove a stronger property : for all $g in G$ and $y in Y$, we have $gyg^{-1} in Y$.
Indeed, if $y=e$ we have $gyg^{-1}=ein Y$. Otherwise, $y$ is of the form $y=(a , b) (c , d)$ (with $a, b, c, d$ numbers from $1$ to $4$ (in some order). We compute that
$$g y g^{-1} = (g(a) , g(b)) (g(c) , g(d))$$
And we check that this is still in $Y$ (note that $Y$ contains all permutations of the form $(a , b) (c , d)$, because for example $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
$endgroup$
add a comment |
$begingroup$
To prove $GY subset YG$, you need to prove that for all $g_1 in G$ and $y_1 in Y$, there exists $y_2 in G$ and $y_2 in Y$ such that $g_1 y_1 = y_2 g_2$, which is equivalent to saying such that $g_1 y_1 g_2^{-1} in Y$. Proving the other inclusion is similar.
We can actually prove a stronger property : for all $g in G$ and $y in Y$, we have $gyg^{-1} in Y$.
Indeed, if $y=e$ we have $gyg^{-1}=ein Y$. Otherwise, $y$ is of the form $y=(a , b) (c , d)$ (with $a, b, c, d$ numbers from $1$ to $4$ (in some order). We compute that
$$g y g^{-1} = (g(a) , g(b)) (g(c) , g(d))$$
And we check that this is still in $Y$ (note that $Y$ contains all permutations of the form $(a , b) (c , d)$, because for example $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
$endgroup$
To prove $GY subset YG$, you need to prove that for all $g_1 in G$ and $y_1 in Y$, there exists $y_2 in G$ and $y_2 in Y$ such that $g_1 y_1 = y_2 g_2$, which is equivalent to saying such that $g_1 y_1 g_2^{-1} in Y$. Proving the other inclusion is similar.
We can actually prove a stronger property : for all $g in G$ and $y in Y$, we have $gyg^{-1} in Y$.
Indeed, if $y=e$ we have $gyg^{-1}=ein Y$. Otherwise, $y$ is of the form $y=(a , b) (c , d)$ (with $a, b, c, d$ numbers from $1$ to $4$ (in some order). We compute that
$$g y g^{-1} = (g(a) , g(b)) (g(c) , g(d))$$
And we check that this is still in $Y$ (note that $Y$ contains all permutations of the form $(a , b) (c , d)$, because for example $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
answered Jan 1 at 17:01
Joel CohenJoel Cohen
7,44412238
7,44412238
add a comment |
add a comment |
2
$begingroup$
And what is $H$?
$endgroup$
– Fakemistake
Jan 1 at 16:32
1
$begingroup$
Hey guys, sorry I copied the question not right. I have edited.
$endgroup$
– TTaJTa4
Jan 1 at 16:36
$begingroup$
$Y$ is a normal subgroup of $S_4$, which means $gY=Yg$ for all $gin S_4$. See the answer by Geoff Robinson.
$endgroup$
– Dietrich Burde
Jan 1 at 16:38
$begingroup$
@TTaJTa4 thank you for fixing the question!
$endgroup$
– layman
Jan 1 at 16:39