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
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$begingroup$
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
1,08416
$endgroup$
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
2,5211717
$endgroup$
$begingroup$
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
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2 Answers
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2 Answers
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active
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$begingroup$
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
1,08416
$endgroup$
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
1,08416
$endgroup$
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
1,08416
$endgroup$
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
1,08416
answered Dec 12 '18 at 18:08
ThinkingThinking
1,08416
answered Dec 12 '18 at 18:08
ThinkingThinking
1,08416
answered Dec 12 '18 at 18:08
ThinkingThinking
1,08416
1,08416
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
2,5211717
$endgroup$
$begingroup$
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
2,5211717
$endgroup$
$begingroup$
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
2,5211717
$endgroup$
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
2,5211717
edited Dec 12 '18 at 22:41
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5211717
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5211717
answered Dec 12 '18 at 18:17
leibnewtzleibnewtz
2,5211717
2,5211717
$begingroup$
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 |
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$begingroup$
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