What are normal closure of a complement of a subgroup and intersection of all normal subgroups?
$begingroup$
What are normal closure of a complement of a subgroup and intersection of all normal subgroups?
If X is a nonempty subset of a group G, define the normal closure X complement of X to be the intersection of all normal subgroups of G that contain X; that is, core H = {g in G | g in aHa^-1 for all a in G} = intersections of {aHa^-1 | a in G}
Sorry first time posting here and I'm not good with codes.
abstract-algebra group-theory
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
$endgroup$
add a comment |
$begingroup$
What are normal closure of a complement of a subgroup and intersection of all normal subgroups?
If X is a nonempty subset of a group G, define the normal closure X complement of X to be the intersection of all normal subgroups of G that contain X; that is, core H = {g in G | g in aHa^-1 for all a in G} = intersections of {aHa^-1 | a in G}
Sorry first time posting here and I'm not good with codes.
abstract-algebra group-theory
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
$endgroup$
add a comment |
$begingroup$
What are normal closure of a complement of a subgroup and intersection of all normal subgroups?
If X is a nonempty subset of a group G, define the normal closure X complement of X to be the intersection of all normal subgroups of G that contain X; that is, core H = {g in G | g in aHa^-1 for all a in G} = intersections of {aHa^-1 | a in G}
Sorry first time posting here and I'm not good with codes.
abstract-algebra group-theory
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
$endgroup$
What are normal closure of a complement of a subgroup and intersection of all normal subgroups?
If X is a nonempty subset of a group G, define the normal closure X complement of X to be the intersection of all normal subgroups of G that contain X; that is, core H = {g in G | g in aHa^-1 for all a in G} = intersections of {aHa^-1 | a in G}
Sorry first time posting here and I'm not good with codes.
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
asked Dec 3 '18 at 19:46
Shawn KooShawn Koo
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
If $H$ is a proper subgroup of $G$, then the subgroup generated by its complement, $langle G-H rangle=G$ (this follows from the fact that a group cannot be the union of two proper subgroups). Hence the normal closure, $(G-H)^G=langle G-H rangle^G=G$.
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
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1 Answer
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$begingroup$
If $H$ is a proper subgroup of $G$, then the subgroup generated by its complement, $langle G-H rangle=G$ (this follows from the fact that a group cannot be the union of two proper subgroups). Hence the normal closure, $(G-H)^G=langle G-H rangle^G=G$.
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
add a comment |
$begingroup$
If $H$ is a proper subgroup of $G$, then the subgroup generated by its complement, $langle G-H rangle=G$ (this follows from the fact that a group cannot be the union of two proper subgroups). Hence the normal closure, $(G-H)^G=langle G-H rangle^G=G$.
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
add a comment |
$begingroup$
If $H$ is a proper subgroup of $G$, then the subgroup generated by its complement, $langle G-H rangle=G$ (this follows from the fact that a group cannot be the union of two proper subgroups). Hence the normal closure, $(G-H)^G=langle G-H rangle^G=G$.
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
If $H$ is a proper subgroup of $G$, then the subgroup generated by its complement, $langle G-H rangle=G$ (this follows from the fact that a group cannot be the union of two proper subgroups). Hence the normal closure, $(G-H)^G=langle G-H rangle^G=G$.
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
answered Dec 3 '18 at 21:10
Nicky HeksterNicky Hekster
28.4k53456
28.4k53456
add a comment |
add a comment |
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