Why is the kernel of $phi : Gmapsto S_m$ included in $Ble G$ where $phi(g)=g:xBmapsto gxB$?
$begingroup$
Let $G$ be a group with $B$ a subgroup of index $m$. There is a homomorphism that associates each element of $G$ to a permutation of the left cosets $xBin G/B$, which is of cardinality $m$.
So these permutations live in $S_m$.
Why is the kernel of that homomorphism included in $B$?
$$kerphi={gin G~:~gxB=xB text{ for all } xin G}=dotsb= bigcaplimits_{gin G}gBg^{-1}$$
I don't see why that is a subset of $B$...
group-theory group-homomorphism
edited Jan 5 at 20:02
SvanN
1,9491422
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group with $B$ a subgroup of index $m$. There is a homomorphism that associates each element of $G$ to a permutation of the left cosets $xBin G/B$, which is of cardinality $m$.
So these permutations live in $S_m$.
Why is the kernel of that homomorphism included in $B$?
$$kerphi={gin G~:~gxB=xB text{ for all } xin G}=dotsb= bigcaplimits_{gin G}gBg^{-1}$$
I don't see why that is a subset of $B$...
group-theory group-homomorphism
edited Jan 5 at 20:02
SvanN
1,9491422
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group with $B$ a subgroup of index $m$. There is a homomorphism that associates each element of $G$ to a permutation of the left cosets $xBin G/B$, which is of cardinality $m$.
So these permutations live in $S_m$.
Why is the kernel of that homomorphism included in $B$?
$$kerphi={gin G~:~gxB=xB text{ for all } xin G}=dotsb= bigcaplimits_{gin G}gBg^{-1}$$
I don't see why that is a subset of $B$...
group-theory group-homomorphism
edited Jan 5 at 20:02
SvanN
1,9491422
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
$endgroup$
Let $G$ be a group with $B$ a subgroup of index $m$. There is a homomorphism that associates each element of $G$ to a permutation of the left cosets $xBin G/B$, which is of cardinality $m$.
So these permutations live in $S_m$.
Why is the kernel of that homomorphism included in $B$?
$$kerphi={gin G~:~gxB=xB text{ for all } xin G}=dotsb= bigcaplimits_{gin G}gBg^{-1}$$
I don't see why that is a subset of $B$...
group-theory group-homomorphism
group-theory group-homomorphism
edited Jan 5 at 20:02
SvanN
1,9491422
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
edited Jan 5 at 20:02
SvanN
1,9491422
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
edited Jan 5 at 20:02
SvanN
1,9491422
edited Jan 5 at 20:02
SvanN
1,9491422
edited Jan 5 at 20:02
SvanN
1,9491422
1,9491422
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
asked Jan 5 at 19:06
John CataldoJohn Cataldo
1,1071216
1,1071216
add a comment |
add a comment |
1 Answer
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$begingroup$
Because take $g=e$ and you will get that $B$ is one of the groups in the intersection. Hence $B$ contains the intersection.
answered Jan 5 at 19:10
MarkMark
6,230416
$endgroup$
add a comment |
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$begingroup$
Because take $g=e$ and you will get that $B$ is one of the groups in the intersection. Hence $B$ contains the intersection.
answered Jan 5 at 19:10
MarkMark
6,230416
$endgroup$
add a comment |
$begingroup$
Because take $g=e$ and you will get that $B$ is one of the groups in the intersection. Hence $B$ contains the intersection.
answered Jan 5 at 19:10
MarkMark
6,230416
$endgroup$
add a comment |
$begingroup$
Because take $g=e$ and you will get that $B$ is one of the groups in the intersection. Hence $B$ contains the intersection.
answered Jan 5 at 19:10
MarkMark
6,230416
$endgroup$
Because take $g=e$ and you will get that $B$ is one of the groups in the intersection. Hence $B$ contains the intersection.
answered Jan 5 at 19:10
MarkMark
6,230416
answered Jan 5 at 19:10
MarkMark
6,230416
answered Jan 5 at 19:10
MarkMark
6,230416
answered Jan 5 at 19:10
MarkMark
6,230416
6,230416
add a comment |
add a comment |
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