Conjugacy classes in subgroup of index 2
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Let $G$ be a finite group, $H$ a subgroup of index 2. It is well-known how to count the conjugacy classes of $H$ once we know those of $G$: look at which ones get split into two classes, based on whether the centralizer is in $H$ or not.
I was wondering: is there a sort of converse for this statement? If we have a good knowledge of the conjugacy classes of $H$, can we study the conjugacy classes of $G$? What about conjugacy classes of elements of a given order?
What if $H$ is furthermore simple? What if $G$ splits over $H$?
I am particularly interested in the number of conjugacy classes of involutions of $G$ knowing the involution structure of $H$.
abstract-algebra group-theory finite-groups normal-subgroups involutions
asked Dec 5 '17 at 8:31
user404944
738212
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up vote
1
down vote
favorite
Let $G$ be a finite group, $H$ a subgroup of index 2. It is well-known how to count the conjugacy classes of $H$ once we know those of $G$: look at which ones get split into two classes, based on whether the centralizer is in $H$ or not.
I was wondering: is there a sort of converse for this statement? If we have a good knowledge of the conjugacy classes of $H$, can we study the conjugacy classes of $G$? What about conjugacy classes of elements of a given order?
What if $H$ is furthermore simple? What if $G$ splits over $H$?
I am particularly interested in the number of conjugacy classes of involutions of $G$ knowing the involution structure of $H$.
abstract-algebra group-theory finite-groups normal-subgroups involutions
asked Dec 5 '17 at 8:31
user404944
738212
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $G$ be a finite group, $H$ a subgroup of index 2. It is well-known how to count the conjugacy classes of $H$ once we know those of $G$: look at which ones get split into two classes, based on whether the centralizer is in $H$ or not.
I was wondering: is there a sort of converse for this statement? If we have a good knowledge of the conjugacy classes of $H$, can we study the conjugacy classes of $G$? What about conjugacy classes of elements of a given order?
What if $H$ is furthermore simple? What if $G$ splits over $H$?
I am particularly interested in the number of conjugacy classes of involutions of $G$ knowing the involution structure of $H$.
abstract-algebra group-theory finite-groups normal-subgroups involutions
asked Dec 5 '17 at 8:31
user404944
738212
Let $G$ be a finite group, $H$ a subgroup of index 2. It is well-known how to count the conjugacy classes of $H$ once we know those of $G$: look at which ones get split into two classes, based on whether the centralizer is in $H$ or not.
I was wondering: is there a sort of converse for this statement? If we have a good knowledge of the conjugacy classes of $H$, can we study the conjugacy classes of $G$? What about conjugacy classes of elements of a given order?
What if $H$ is furthermore simple? What if $G$ splits over $H$?
I am particularly interested in the number of conjugacy classes of involutions of $G$ knowing the involution structure of $H$.
abstract-algebra group-theory finite-groups normal-subgroups involutions
abstract-algebra group-theory finite-groups normal-subgroups involutions
asked Dec 5 '17 at 8:31
user404944
738212
asked Dec 5 '17 at 8:31
user404944
738212
edited Dec 5 '17 at 16:39
asked Dec 5 '17 at 8:31
user404944
738212
asked Dec 5 '17 at 8:31
user404944
738212
asked Dec 5 '17 at 8:31
user404944
738212
738212
add a comment |
add a comment |
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