Intersection of cosets from possibly distinct subgroups is either empty or a coset of the intersection...
$begingroup$
Let $G$ be a group and $H leq G, K leq G$.
The intersection of two left cosets (one from $H$ and the other from $K$) is either empty or a coset of
$Hcap K$
I'm having some trouble proving this in a good way.
We know that, if $G$ is a group and $H leq G, K leq G$ then $(Hcap K) leq G$.
Let us a consider $Hcap K$ and some coset $a(Hcap K)$ of $Hcap K$ for a arbitrary $ain G$
If $x in a(Hcap K) implies x =ab,$ for some $b in Hcap K implies x in aH$ and $x in aK implies x in aHcap aK $. But this is not what I want, right? Or can this implication be done the other way as well?
One thought is to split it up in to cases.
- $Hcap K = {e }$, where e is the identity.
- $H = K$ or $H subset K$ or $K subset H$ or $Hcap K neq {e }$.
But I'm not sure if this would get me anywhere?
If I want to prove that the intersection must be empty I'm not sure where to start. I want to assume that some $xin G$ is contained in the intersection of two cosets and then perhaps derive that it must be empty.
Should I then take $x in aHcap aK $ or $x in aHcap bK $ where $a,b$ may or may not be distinct?
If I take $x in aHcap aK $ then $x = ah, hin H$ and $x = ak, kin K implies x = ah = ak implies h = k$. And I'm stuck.
If I take $x in aHcap bK $ then $x = ah, hin H$ and $x = bk, kin K implies x = ah = bk implies a^{-1}bk in H$ And I'm also stuck here.
Any hints on how to solve this?
abstract-algebra group-theory
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group and $H leq G, K leq G$.
The intersection of two left cosets (one from $H$ and the other from $K$) is either empty or a coset of
$Hcap K$
I'm having some trouble proving this in a good way.
We know that, if $G$ is a group and $H leq G, K leq G$ then $(Hcap K) leq G$.
Let us a consider $Hcap K$ and some coset $a(Hcap K)$ of $Hcap K$ for a arbitrary $ain G$
If $x in a(Hcap K) implies x =ab,$ for some $b in Hcap K implies x in aH$ and $x in aK implies x in aHcap aK $. But this is not what I want, right? Or can this implication be done the other way as well?
One thought is to split it up in to cases.
- $Hcap K = {e }$, where e is the identity.
- $H = K$ or $H subset K$ or $K subset H$ or $Hcap K neq {e }$.
But I'm not sure if this would get me anywhere?
If I want to prove that the intersection must be empty I'm not sure where to start. I want to assume that some $xin G$ is contained in the intersection of two cosets and then perhaps derive that it must be empty.
Should I then take $x in aHcap aK $ or $x in aHcap bK $ where $a,b$ may or may not be distinct?
If I take $x in aHcap aK $ then $x = ah, hin H$ and $x = ak, kin K implies x = ah = ak implies h = k$. And I'm stuck.
If I take $x in aHcap bK $ then $x = ah, hin H$ and $x = bk, kin K implies x = ah = bk implies a^{-1}bk in H$ And I'm also stuck here.
Any hints on how to solve this?
abstract-algebra group-theory
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
$endgroup$
$begingroup$
Ooooops. This is a duplicate of math.stackexchange.com/questions/270703/…
$endgroup$
– John Smith
Feb 23 '14 at 18:12
add a comment |
$begingroup$
Let $G$ be a group and $H leq G, K leq G$.
The intersection of two left cosets (one from $H$ and the other from $K$) is either empty or a coset of
$Hcap K$
I'm having some trouble proving this in a good way.
We know that, if $G$ is a group and $H leq G, K leq G$ then $(Hcap K) leq G$.
Let us a consider $Hcap K$ and some coset $a(Hcap K)$ of $Hcap K$ for a arbitrary $ain G$
If $x in a(Hcap K) implies x =ab,$ for some $b in Hcap K implies x in aH$ and $x in aK implies x in aHcap aK $. But this is not what I want, right? Or can this implication be done the other way as well?
One thought is to split it up in to cases.
- $Hcap K = {e }$, where e is the identity.
- $H = K$ or $H subset K$ or $K subset H$ or $Hcap K neq {e }$.
But I'm not sure if this would get me anywhere?
If I want to prove that the intersection must be empty I'm not sure where to start. I want to assume that some $xin G$ is contained in the intersection of two cosets and then perhaps derive that it must be empty.
Should I then take $x in aHcap aK $ or $x in aHcap bK $ where $a,b$ may or may not be distinct?
If I take $x in aHcap aK $ then $x = ah, hin H$ and $x = ak, kin K implies x = ah = ak implies h = k$. And I'm stuck.
If I take $x in aHcap bK $ then $x = ah, hin H$ and $x = bk, kin K implies x = ah = bk implies a^{-1}bk in H$ And I'm also stuck here.
Any hints on how to solve this?
abstract-algebra group-theory
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
$endgroup$
Let $G$ be a group and $H leq G, K leq G$.
The intersection of two left cosets (one from $H$ and the other from $K$) is either empty or a coset of
$Hcap K$
I'm having some trouble proving this in a good way.
We know that, if $G$ is a group and $H leq G, K leq G$ then $(Hcap K) leq G$.
Let us a consider $Hcap K$ and some coset $a(Hcap K)$ of $Hcap K$ for a arbitrary $ain G$
If $x in a(Hcap K) implies x =ab,$ for some $b in Hcap K implies x in aH$ and $x in aK implies x in aHcap aK $. But this is not what I want, right? Or can this implication be done the other way as well?
One thought is to split it up in to cases.
- $Hcap K = {e }$, where e is the identity.
- $H = K$ or $H subset K$ or $K subset H$ or $Hcap K neq {e }$.
But I'm not sure if this would get me anywhere?
If I want to prove that the intersection must be empty I'm not sure where to start. I want to assume that some $xin G$ is contained in the intersection of two cosets and then perhaps derive that it must be empty.
Should I then take $x in aHcap aK $ or $x in aHcap bK $ where $a,b$ may or may not be distinct?
If I take $x in aHcap aK $ then $x = ah, hin H$ and $x = ak, kin K implies x = ah = ak implies h = k$. And I'm stuck.
If I take $x in aHcap bK $ then $x = ah, hin H$ and $x = bk, kin K implies x = ah = bk implies a^{-1}bk in H$ And I'm also stuck here.
Any hints on how to solve this?
abstract-algebra group-theory
abstract-algebra group-theory
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
asked Feb 23 '14 at 12:25
John SmithJohn Smith
952826
952826
$begingroup$
Ooooops. This is a duplicate of math.stackexchange.com/questions/270703/…
$endgroup$
– John Smith
Feb 23 '14 at 18:12
add a comment |
$begingroup$
Ooooops. This is a duplicate of math.stackexchange.com/questions/270703/…
$endgroup$
– John Smith
Feb 23 '14 at 18:12
$begingroup$
Ooooops. This is a duplicate of math.stackexchange.com/questions/270703/…
$endgroup$
– John Smith
Feb 23 '14 at 18:12
$begingroup$
Ooooops. This is a duplicate of math.stackexchange.com/questions/270703/…
$endgroup$
– John Smith
Feb 23 '14 at 18:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$a$ and $b$ need not to be equal. You have to show that the intersection between a left coset $aH$ of $H$ and a left coset $bK$ of $K$ is a left coset of $H cap K$, that is $aH cap bK = c(H cap K)$ for some $c in G$.
As you hinted, suppose $aH cap bK neq emptyset$ and let $x in aH cap bK$. Then $x = ah = bk$ for some $h in H$, $k in K$ and $h = xa^{-1}$, $k = xb^{-1}$.
We shall now prove that $aH = xH$. Let $g in aH$, then $$g = ah' = xh^{-1}h' in xH$$
for some $h' in H$.
Conversely, if $g in xH$, then $$g = xh' = ahh' in aH$$ for some $h' in H$.
Similarly, we have $bK = xK$.
Hence (explain why!) $aH cap bK = xH cap xK = x(H cap K)$.
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
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add a comment |
$begingroup$
dani_s's answer has a typo in the second paragraph. It should state $a=xh^{-1}, b=xk^{-1}$ instead ($h=xa^{-1}$ would be wrong without assuming abelian group).
Also the middle section of the proof can be shortened:
$$aH=xh^{-1}H=x(h^{-1}H)=xH quadtext{ since }h^{-1} in H$$
and
$$bK=xk^{-1}K=x(k^{-1}K)=xK quadtext{ since }k^{-1} in K.$$
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
$endgroup$
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– idea
Jan 20 at 11:33
$begingroup$
That is your system's design problem, is it not?
$endgroup$
– Xiaohai Zhang
Jan 21 at 0:40
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$a$ and $b$ need not to be equal. You have to show that the intersection between a left coset $aH$ of $H$ and a left coset $bK$ of $K$ is a left coset of $H cap K$, that is $aH cap bK = c(H cap K)$ for some $c in G$.
As you hinted, suppose $aH cap bK neq emptyset$ and let $x in aH cap bK$. Then $x = ah = bk$ for some $h in H$, $k in K$ and $h = xa^{-1}$, $k = xb^{-1}$.
We shall now prove that $aH = xH$. Let $g in aH$, then $$g = ah' = xh^{-1}h' in xH$$
for some $h' in H$.
Conversely, if $g in xH$, then $$g = xh' = ahh' in aH$$ for some $h' in H$.
Similarly, we have $bK = xK$.
Hence (explain why!) $aH cap bK = xH cap xK = x(H cap K)$.
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
$endgroup$
add a comment |
$begingroup$
$a$ and $b$ need not to be equal. You have to show that the intersection between a left coset $aH$ of $H$ and a left coset $bK$ of $K$ is a left coset of $H cap K$, that is $aH cap bK = c(H cap K)$ for some $c in G$.
As you hinted, suppose $aH cap bK neq emptyset$ and let $x in aH cap bK$. Then $x = ah = bk$ for some $h in H$, $k in K$ and $h = xa^{-1}$, $k = xb^{-1}$.
We shall now prove that $aH = xH$. Let $g in aH$, then $$g = ah' = xh^{-1}h' in xH$$
for some $h' in H$.
Conversely, if $g in xH$, then $$g = xh' = ahh' in aH$$ for some $h' in H$.
Similarly, we have $bK = xK$.
Hence (explain why!) $aH cap bK = xH cap xK = x(H cap K)$.
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
$endgroup$
add a comment |
$begingroup$
$a$ and $b$ need not to be equal. You have to show that the intersection between a left coset $aH$ of $H$ and a left coset $bK$ of $K$ is a left coset of $H cap K$, that is $aH cap bK = c(H cap K)$ for some $c in G$.
As you hinted, suppose $aH cap bK neq emptyset$ and let $x in aH cap bK$. Then $x = ah = bk$ for some $h in H$, $k in K$ and $h = xa^{-1}$, $k = xb^{-1}$.
We shall now prove that $aH = xH$. Let $g in aH$, then $$g = ah' = xh^{-1}h' in xH$$
for some $h' in H$.
Conversely, if $g in xH$, then $$g = xh' = ahh' in aH$$ for some $h' in H$.
Similarly, we have $bK = xK$.
Hence (explain why!) $aH cap bK = xH cap xK = x(H cap K)$.
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
$endgroup$
$a$ and $b$ need not to be equal. You have to show that the intersection between a left coset $aH$ of $H$ and a left coset $bK$ of $K$ is a left coset of $H cap K$, that is $aH cap bK = c(H cap K)$ for some $c in G$.
As you hinted, suppose $aH cap bK neq emptyset$ and let $x in aH cap bK$. Then $x = ah = bk$ for some $h in H$, $k in K$ and $h = xa^{-1}$, $k = xb^{-1}$.
We shall now prove that $aH = xH$. Let $g in aH$, then $$g = ah' = xh^{-1}h' in xH$$
for some $h' in H$.
Conversely, if $g in xH$, then $$g = xh' = ahh' in aH$$ for some $h' in H$.
Similarly, we have $bK = xK$.
Hence (explain why!) $aH cap bK = xH cap xK = x(H cap K)$.
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
answered Feb 23 '14 at 13:41
dani_sdani_s
1,434711
1,434711
add a comment |
add a comment |
$begingroup$
dani_s's answer has a typo in the second paragraph. It should state $a=xh^{-1}, b=xk^{-1}$ instead ($h=xa^{-1}$ would be wrong without assuming abelian group).
Also the middle section of the proof can be shortened:
$$aH=xh^{-1}H=x(h^{-1}H)=xH quadtext{ since }h^{-1} in H$$
and
$$bK=xk^{-1}K=x(k^{-1}K)=xK quadtext{ since }k^{-1} in K.$$
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
$endgroup$
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– idea
Jan 20 at 11:33
$begingroup$
That is your system's design problem, is it not?
$endgroup$
– Xiaohai Zhang
Jan 21 at 0:40
add a comment |
$begingroup$
dani_s's answer has a typo in the second paragraph. It should state $a=xh^{-1}, b=xk^{-1}$ instead ($h=xa^{-1}$ would be wrong without assuming abelian group).
Also the middle section of the proof can be shortened:
$$aH=xh^{-1}H=x(h^{-1}H)=xH quadtext{ since }h^{-1} in H$$
and
$$bK=xk^{-1}K=x(k^{-1}K)=xK quadtext{ since }k^{-1} in K.$$
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
$endgroup$
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– idea
Jan 20 at 11:33
$begingroup$
That is your system's design problem, is it not?
$endgroup$
– Xiaohai Zhang
Jan 21 at 0:40
add a comment |
$begingroup$
dani_s's answer has a typo in the second paragraph. It should state $a=xh^{-1}, b=xk^{-1}$ instead ($h=xa^{-1}$ would be wrong without assuming abelian group).
Also the middle section of the proof can be shortened:
$$aH=xh^{-1}H=x(h^{-1}H)=xH quadtext{ since }h^{-1} in H$$
and
$$bK=xk^{-1}K=x(k^{-1}K)=xK quadtext{ since }k^{-1} in K.$$
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
$endgroup$
dani_s's answer has a typo in the second paragraph. It should state $a=xh^{-1}, b=xk^{-1}$ instead ($h=xa^{-1}$ would be wrong without assuming abelian group).
Also the middle section of the proof can be shortened:
$$aH=xh^{-1}H=x(h^{-1}H)=xH quadtext{ since }h^{-1} in H$$
and
$$bK=xk^{-1}K=x(k^{-1}K)=xK quadtext{ since }k^{-1} in K.$$
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
answered Jan 20 at 9:56
Xiaohai ZhangXiaohai Zhang
212
212
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– idea
Jan 20 at 11:33
$begingroup$
That is your system's design problem, is it not?
$endgroup$
– Xiaohai Zhang
Jan 21 at 0:40
add a comment |
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– idea
Jan 20 at 11:33
$begingroup$
That is your system's design problem, is it not?
$endgroup$
– Xiaohai Zhang
Jan 21 at 0:40
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– idea
Jan 20 at 11:33
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– idea
Jan 20 at 11:33
$begingroup$
That is your system's design problem, is it not?
$endgroup$
– Xiaohai Zhang
Jan 21 at 0:40
$begingroup$
That is your system's design problem, is it not?
$endgroup$
– Xiaohai Zhang
Jan 21 at 0:40
add a comment |
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$begingroup$
Ooooops. This is a duplicate of math.stackexchange.com/questions/270703/…
$endgroup$
– John Smith
Feb 23 '14 at 18:12