Proving ${gY:gin X}={Yg:gin X}$ [duplicate]












0












$begingroup$



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:



$${gY:gin X}={Yg:gin X}$$










share|cite|improve this question











$endgroup$



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.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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
















0












$begingroup$



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:



$${gY:gin X}={Yg:gin X}$$










share|cite|improve this question











$endgroup$



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.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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














0












0








0





$begingroup$



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:



$${gY:gin X}={Yg:gin X}$$










share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 16:35







TTaJTa4

















asked Jan 1 at 16:30









TTaJTa4TTaJTa4

1665




1665




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.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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
Users with the  group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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














  • 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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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 $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).






        share|cite|improve this answer









        $endgroup$



        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 $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=ldots$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 17:01









        Joel CohenJoel Cohen

        7,33412137




        7,33412137















            Popular posts from this blog

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei