Commutator group normal in G [duplicate]
$begingroup$
This question already has an answer here:
How to show that the commutator subgroup is a normal subgroup
1 answer
$G$ is a group with $e$ as it's identity element.
The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$
Show that $[G,G]triangleleft G$
Solution:
If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].
$$g^{-1}[x,y]g$$
$$g^{-1}x^{-1}y^{-1}xyg$$
I am stuck here. I am not sure how to proceed since we have three elements...
Thanks for the help!
group-theory
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
$endgroup$
marked as duplicate by Arturo Magidin 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();
}
);
});
});
Dec 8 '18 at 3:54
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 the commutator subgroup is a normal subgroup
1 answer
$G$ is a group with $e$ as it's identity element.
The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$
Show that $[G,G]triangleleft G$
Solution:
If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].
$$g^{-1}[x,y]g$$
$$g^{-1}x^{-1}y^{-1}xyg$$
I am stuck here. I am not sure how to proceed since we have three elements...
Thanks for the help!
group-theory
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
$endgroup$
marked as duplicate by Arturo Magidin 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();
}
);
});
});
Dec 8 '18 at 3:54
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 the commutator subgroup is a normal subgroup
1 answer
$G$ is a group with $e$ as it's identity element.
The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$
Show that $[G,G]triangleleft G$
Solution:
If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].
$$g^{-1}[x,y]g$$
$$g^{-1}x^{-1}y^{-1}xyg$$
I am stuck here. I am not sure how to proceed since we have three elements...
Thanks for the help!
group-theory
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
$endgroup$
This question already has an answer here:
How to show that the commutator subgroup is a normal subgroup
1 answer
$G$ is a group with $e$ as it's identity element.
The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$
Show that $[G,G]triangleleft G$
Solution:
If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].
$$g^{-1}[x,y]g$$
$$g^{-1}x^{-1}y^{-1}xyg$$
I am stuck here. I am not sure how to proceed since we have three elements...
Thanks for the help!
This question already has an answer here:
How to show that the commutator subgroup is a normal subgroup
1 answer
group-theory
group-theory
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
asked Dec 8 '18 at 3:32
JoeyFJoeyF
62
62
marked as duplicate by Arturo Magidin 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();
}
);
});
});
Dec 8 '18 at 3:54
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 Arturo Magidin 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();
}
);
});
});
Dec 8 '18 at 3:54
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 |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT :
$$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$
Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
1
$begingroup$
Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:55
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT :
$$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$
Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
1
$begingroup$
Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:55
add a comment |
$begingroup$
HINT :
$$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$
Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
1
$begingroup$
Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:55
add a comment |
$begingroup$
HINT :
$$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$
Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
$endgroup$
HINT :
$$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$
Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
answered Dec 8 '18 at 3:44
Thomas ShelbyThomas Shelby
2,498221
2,498221
1
$begingroup$
Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:55
add a comment |
1
$begingroup$
Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:55
1
1
$begingroup$
Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:55
$begingroup$
Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
$endgroup$
– Arturo Magidin
Dec 8 '18 at 3:55
add a comment |
