Quotient group $G/H$ is abelian iff $[G,G] subseteq H$
$begingroup$
So I've been working on this problem for a while. I just can't seem to find the solution. Also in the previous subquestion, I showed that $G$ is abelian iff $[G,G] = {e}$. Where $e$ is the neutral element of $G$ and $[G,G]$ is the commutator group of $G$.
Let $Htriangleleft G$, Show that $G/H$ is abelian iif $[G,G] subseteq H$.
I've proved this direction
$Rightarrow $)
If $G/H$ is abelian, $forall x,y in G$,
$xHyH = xyH = yxH = yHxH$.
Therefore have that $G$ is abelian. As shown earlier, $[G,G] = {e}$ if $G$ is abelian. So $[G,G] = {e} subseteq H$.
I can't seem to figure out the other direction...
$Leftarrow$)
I only have that for $x,y in G$,
$x^{-1}y^{-1}xy in H$ and that $Htriangleleft G$. It does not seem to lead me anywhere.
Thanks for the help!
group-theory
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
$endgroup$
add a comment |
$begingroup$
So I've been working on this problem for a while. I just can't seem to find the solution. Also in the previous subquestion, I showed that $G$ is abelian iff $[G,G] = {e}$. Where $e$ is the neutral element of $G$ and $[G,G]$ is the commutator group of $G$.
Let $Htriangleleft G$, Show that $G/H$ is abelian iif $[G,G] subseteq H$.
I've proved this direction
$Rightarrow $)
If $G/H$ is abelian, $forall x,y in G$,
$xHyH = xyH = yxH = yHxH$.
Therefore have that $G$ is abelian. As shown earlier, $[G,G] = {e}$ if $G$ is abelian. So $[G,G] = {e} subseteq H$.
I can't seem to figure out the other direction...
$Leftarrow$)
I only have that for $x,y in G$,
$x^{-1}y^{-1}xy in H$ and that $Htriangleleft G$. It does not seem to lead me anywhere.
Thanks for the help!
group-theory
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
$endgroup$
$begingroup$
If $phi $ is a (surjective) morphism $G to A$ with $A$ abelian then $[G,G] subset ker(phi)$. Conversely if $[G,G] subset ker(phi)$ then $phi : G to G/[G,G]to A$ so $A$ is abelian.
$endgroup$
– reuns
Dec 12 '18 at 17:58
add a comment |
$begingroup$
So I've been working on this problem for a while. I just can't seem to find the solution. Also in the previous subquestion, I showed that $G$ is abelian iff $[G,G] = {e}$. Where $e$ is the neutral element of $G$ and $[G,G]$ is the commutator group of $G$.
Let $Htriangleleft G$, Show that $G/H$ is abelian iif $[G,G] subseteq H$.
I've proved this direction
$Rightarrow $)
If $G/H$ is abelian, $forall x,y in G$,
$xHyH = xyH = yxH = yHxH$.
Therefore have that $G$ is abelian. As shown earlier, $[G,G] = {e}$ if $G$ is abelian. So $[G,G] = {e} subseteq H$.
I can't seem to figure out the other direction...
$Leftarrow$)
I only have that for $x,y in G$,
$x^{-1}y^{-1}xy in H$ and that $Htriangleleft G$. It does not seem to lead me anywhere.
Thanks for the help!
group-theory
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
$endgroup$
So I've been working on this problem for a while. I just can't seem to find the solution. Also in the previous subquestion, I showed that $G$ is abelian iff $[G,G] = {e}$. Where $e$ is the neutral element of $G$ and $[G,G]$ is the commutator group of $G$.
Let $Htriangleleft G$, Show that $G/H$ is abelian iif $[G,G] subseteq H$.
I've proved this direction
$Rightarrow $)
If $G/H$ is abelian, $forall x,y in G$,
$xHyH = xyH = yxH = yHxH$.
Therefore have that $G$ is abelian. As shown earlier, $[G,G] = {e}$ if $G$ is abelian. So $[G,G] = {e} subseteq H$.
I can't seem to figure out the other direction...
$Leftarrow$)
I only have that for $x,y in G$,
$x^{-1}y^{-1}xy in H$ and that $Htriangleleft G$. It does not seem to lead me anywhere.
Thanks for the help!
group-theory
group-theory
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
asked Dec 12 '18 at 17:56
JoeyFJoeyF
62
62
$begingroup$
If $phi $ is a (surjective) morphism $G to A$ with $A$ abelian then $[G,G] subset ker(phi)$. Conversely if $[G,G] subset ker(phi)$ then $phi : G to G/[G,G]to A$ so $A$ is abelian.
$endgroup$
– reuns
Dec 12 '18 at 17:58
add a comment |
$begingroup$
If $phi $ is a (surjective) morphism $G to A$ with $A$ abelian then $[G,G] subset ker(phi)$. Conversely if $[G,G] subset ker(phi)$ then $phi : G to G/[G,G]to A$ so $A$ is abelian.
$endgroup$
– reuns
Dec 12 '18 at 17:58
$begingroup$
If $phi $ is a (surjective) morphism $G to A$ with $A$ abelian then $[G,G] subset ker(phi)$. Conversely if $[G,G] subset ker(phi)$ then $phi : G to G/[G,G]to A$ so $A$ is abelian.
$endgroup$
– reuns
Dec 12 '18 at 17:58
$begingroup$
If $phi $ is a (surjective) morphism $G to A$ with $A$ abelian then $[G,G] subset ker(phi)$. Conversely if $[G,G] subset ker(phi)$ then $phi : G to G/[G,G]to A$ so $A$ is abelian.
$endgroup$
– reuns
Dec 12 '18 at 17:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let :
$$pi : G to G / H$$
Then :
$$forall x, y in G, pi([x,y]) = [pi(x), pi(y)]$$
Yet since $[G,G] subset H$ then $pi([G,G]) = {e}$. Hence we have :
$$forall x, y, [pi(x), pi(y)] = e$$
So : $G/H$ is abelian.
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
$endgroup$
add a comment |
$begingroup$
Recall that $[G,G]=langle xyx^{-1}y^{-1} | x,y in Grangle$.
If $G/H$ is abelian, this means $xyH=yxH$, for all $x,y in G$. Hence $xyx^{-1}y^{-1} in H$, so $[G,G] subseteq H$. Be careful here, we don't know that $G$ is abelian as you suggest.
On the other hand, if $[G,G] subseteq H$, then for all $xyx^{-1}y^{-1} in [G,G]$ we have $xyx^{-1}y^{-1}H=H$. But this means $xyH=yxH$, so $G/H$ is abelian.
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
$endgroup$
$begingroup$
Beware the commutators are not a subgroup. They only *generate $G,G]$.
$endgroup$
– Bernard
Dec 12 '18 at 18:37
$begingroup$
For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem
$endgroup$
– leibnewtz
Dec 12 '18 at 22:40
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037015%2fquotient-group-g-h-is-abelian-iff-g-g-subseteq-h%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let :
$$pi : G to G / H$$
Then :
$$forall x, y in G, pi([x,y]) = [pi(x), pi(y)]$$
Yet since $[G,G] subset H$ then $pi([G,G]) = {e}$. Hence we have :
$$forall x, y, [pi(x), pi(y)] = e$$
So : $G/H$ is abelian.
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
$endgroup$
add a comment |
$begingroup$
Let :
$$pi : G to G / H$$
Then :
$$forall x, y in G, pi([x,y]) = [pi(x), pi(y)]$$
Yet since $[G,G] subset H$ then $pi([G,G]) = {e}$. Hence we have :
$$forall x, y, [pi(x), pi(y)] = e$$
So : $G/H$ is abelian.
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
$endgroup$
add a comment |
$begingroup$
Let :
$$pi : G to G / H$$
Then :
$$forall x, y in G, pi([x,y]) = [pi(x), pi(y)]$$
Yet since $[G,G] subset H$ then $pi([G,G]) = {e}$. Hence we have :
$$forall x, y, [pi(x), pi(y)] = e$$
So : $G/H$ is abelian.
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
$endgroup$
Let :
$$pi : G to G / H$$
Then :
$$forall x, y in G, pi([x,y]) = [pi(x), pi(y)]$$
Yet since $[G,G] subset H$ then $pi([G,G]) = {e}$. Hence we have :
$$forall x, y, [pi(x), pi(y)] = e$$
So : $G/H$ is abelian.
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
answered Dec 12 '18 at 18:08
ThinkingThinking
1,13916
1,13916
add a comment |
add a comment |
$begingroup$
Recall that $[G,G]=langle xyx^{-1}y^{-1} | x,y in Grangle$.
If $G/H$ is abelian, this means $xyH=yxH$, for all $x,y in G$. Hence $xyx^{-1}y^{-1} in H$, so $[G,G] subseteq H$. Be careful here, we don't know that $G$ is abelian as you suggest.
On the other hand, if $[G,G] subseteq H$, then for all $xyx^{-1}y^{-1} in [G,G]$ we have $xyx^{-1}y^{-1}H=H$. But this means $xyH=yxH$, so $G/H$ is abelian.
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
$endgroup$
$begingroup$
Beware the commutators are not a subgroup. They only *generate $G,G]$.
$endgroup$
– Bernard
Dec 12 '18 at 18:37
$begingroup$
For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem
$endgroup$
– leibnewtz
Dec 12 '18 at 22:40
add a comment |
$begingroup$
Recall that $[G,G]=langle xyx^{-1}y^{-1} | x,y in Grangle$.
If $G/H$ is abelian, this means $xyH=yxH$, for all $x,y in G$. Hence $xyx^{-1}y^{-1} in H$, so $[G,G] subseteq H$. Be careful here, we don't know that $G$ is abelian as you suggest.
On the other hand, if $[G,G] subseteq H$, then for all $xyx^{-1}y^{-1} in [G,G]$ we have $xyx^{-1}y^{-1}H=H$. But this means $xyH=yxH$, so $G/H$ is abelian.
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
$endgroup$
$begingroup$
Beware the commutators are not a subgroup. They only *generate $G,G]$.
$endgroup$
– Bernard
Dec 12 '18 at 18:37
$begingroup$
For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem
$endgroup$
– leibnewtz
Dec 12 '18 at 22:40
add a comment |
$begingroup$
Recall that $[G,G]=langle xyx^{-1}y^{-1} | x,y in Grangle$.
If $G/H$ is abelian, this means $xyH=yxH$, for all $x,y in G$. Hence $xyx^{-1}y^{-1} in H$, so $[G,G] subseteq H$. Be careful here, we don't know that $G$ is abelian as you suggest.
On the other hand, if $[G,G] subseteq H$, then for all $xyx^{-1}y^{-1} in [G,G]$ we have $xyx^{-1}y^{-1}H=H$. But this means $xyH=yxH$, so $G/H$ is abelian.
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
$endgroup$
Recall that $[G,G]=langle xyx^{-1}y^{-1} | x,y in Grangle$.
If $G/H$ is abelian, this means $xyH=yxH$, for all $x,y in G$. Hence $xyx^{-1}y^{-1} in H$, so $[G,G] subseteq H$. Be careful here, we don't know that $G$ is abelian as you suggest.
On the other hand, if $[G,G] subseteq H$, then for all $xyx^{-1}y^{-1} in [G,G]$ we have $xyx^{-1}y^{-1}H=H$. But this means $xyH=yxH$, so $G/H$ is abelian.
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
edited Dec 12 '18 at 22:41
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5861717
2,5861717
$begingroup$
Beware the commutators are not a subgroup. They only *generate $G,G]$.
$endgroup$
– Bernard
Dec 12 '18 at 18:37
$begingroup$
For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem
$endgroup$
– leibnewtz
Dec 12 '18 at 22:40
add a comment |
$begingroup$
Beware the commutators are not a subgroup. They only *generate $G,G]$.
$endgroup$
– Bernard
Dec 12 '18 at 18:37
$begingroup$
For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem
$endgroup$
– leibnewtz
Dec 12 '18 at 22:40
$begingroup$
Beware the commutators are not a subgroup. They only *generate $G,G]$.
$endgroup$
– Bernard
Dec 12 '18 at 18:37
$begingroup$
Beware the commutators are not a subgroup. They only *generate $G,G]$.
$endgroup$
– Bernard
Dec 12 '18 at 18:37
$begingroup$
For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem
$endgroup$
– leibnewtz
Dec 12 '18 at 22:40
$begingroup$
For the purposes of this proof it makes no difference, but I changed to brackets instead of curly braces if that was the problem
$endgroup$
– leibnewtz
Dec 12 '18 at 22:40
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037015%2fquotient-group-g-h-is-abelian-iff-g-g-subseteq-h%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
If $phi $ is a (surjective) morphism $G to A$ with $A$ abelian then $[G,G] subset ker(phi)$. Conversely if $[G,G] subset ker(phi)$ then $phi : G to G/[G,G]to A$ so $A$ is abelian.
$endgroup$
– reuns
Dec 12 '18 at 17:58