Show this equality : $[G: Psi(H) ] = [G:H] $ [closed]
1
$begingroup$
$G$ group, $Psi in text{Aut}(G) $
Show that:
If H is of index (i mean that the cardinal of $G$ over $H$ is finite)
then : $$[G: Psi(H) ] = [G:H] $$
group-theory automorphism-group
edited Jan 15 at 18:08
Surb
38.4k94478
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
$endgroup$
closed as off-topic by Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin Jan 16 at 10:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 4 more comments
1
$begingroup$
$G$ group, $Psi in text{Aut}(G) $
Show that:
If H is of index (i mean that the cardinal of $G$ over $H$ is finite)
then : $$[G: Psi(H) ] = [G:H] $$
group-theory automorphism-group
edited Jan 15 at 18:08
Surb
38.4k94478
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
$endgroup$
closed as off-topic by Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin Jan 16 at 10:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@Surb No, that is not true.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:15
$begingroup$
@Surb Right, but subgroups of the same order can have different index in general if the group is not finite.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:19
$begingroup$
So any idea for this exercice ?
$endgroup$
– Marine Galantin
Jan 15 at 18:51
$begingroup$
Can't you just use the automorphism to establish a bijection between the cosets?
$endgroup$
– verret
Jan 15 at 18:55
$begingroup$
in fact that's what I want to do, but I'm not sure if this is true : in fact if I write an attempt for this exercice, I would just have written that there is a bijection between the two sets and this yields the result. Maybe there are a few things to add in between ?
$endgroup$
– Marine Galantin
Jan 15 at 19:02
|
show 4 more comments
1
1
1
$begingroup$
$G$ group, $Psi in text{Aut}(G) $
Show that:
If H is of index (i mean that the cardinal of $G$ over $H$ is finite)
then : $$[G: Psi(H) ] = [G:H] $$
group-theory automorphism-group
edited Jan 15 at 18:08
Surb
38.4k94478
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
$endgroup$
$G$ group, $Psi in text{Aut}(G) $
Show that:
If H is of index (i mean that the cardinal of $G$ over $H$ is finite)
then : $$[G: Psi(H) ] = [G:H] $$
group-theory automorphism-group
group-theory automorphism-group
edited Jan 15 at 18:08
Surb
38.4k94478
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
edited Jan 15 at 18:08
Surb
38.4k94478
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
edited Jan 15 at 18:08
Surb
38.4k94478
edited Jan 15 at 18:08
Surb
38.4k94478
edited Jan 15 at 18:08
Surb
38.4k94478
38.4k94478
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
asked Jan 15 at 18:02
Marine GalantinMarine Galantin
945419
945419
closed as off-topic by Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin Jan 16 at 10:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin Jan 16 at 10:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Cesareo, A. Pongrácz, José Carlos Santos, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@Surb No, that is not true.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:15
$begingroup$
@Surb Right, but subgroups of the same order can have different index in general if the group is not finite.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:19
$begingroup$
So any idea for this exercice ?
$endgroup$
– Marine Galantin
Jan 15 at 18:51
$begingroup$
Can't you just use the automorphism to establish a bijection between the cosets?
$endgroup$
– verret
Jan 15 at 18:55
$begingroup$
in fact that's what I want to do, but I'm not sure if this is true : in fact if I write an attempt for this exercice, I would just have written that there is a bijection between the two sets and this yields the result. Maybe there are a few things to add in between ?
$endgroup$
– Marine Galantin
Jan 15 at 19:02
|
show 4 more comments
$begingroup$
@Surb No, that is not true.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:15
$begingroup$
@Surb Right, but subgroups of the same order can have different index in general if the group is not finite.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:19
$begingroup$
So any idea for this exercice ?
$endgroup$
– Marine Galantin
Jan 15 at 18:51
$begingroup$
Can't you just use the automorphism to establish a bijection between the cosets?
$endgroup$
– verret
Jan 15 at 18:55
$begingroup$
in fact that's what I want to do, but I'm not sure if this is true : in fact if I write an attempt for this exercice, I would just have written that there is a bijection between the two sets and this yields the result. Maybe there are a few things to add in between ?
$endgroup$
– Marine Galantin
Jan 15 at 19:02
$begingroup$
@Surb No, that is not true.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:15
$begingroup$
@Surb No, that is not true.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:15
$begingroup$
@Surb Right, but subgroups of the same order can have different index in general if the group is not finite.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:19
$begingroup$
@Surb Right, but subgroups of the same order can have different index in general if the group is not finite.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:19
$begingroup$
So any idea for this exercice ?
$endgroup$
– Marine Galantin
Jan 15 at 18:51
$begingroup$
So any idea for this exercice ?
$endgroup$
– Marine Galantin
Jan 15 at 18:51
$begingroup$
Can't you just use the automorphism to establish a bijection between the cosets?
$endgroup$
– verret
Jan 15 at 18:55
$begingroup$
Can't you just use the automorphism to establish a bijection between the cosets?
$endgroup$
– verret
Jan 15 at 18:55
$begingroup$
in fact that's what I want to do, but I'm not sure if this is true : in fact if I write an attempt for this exercice, I would just have written that there is a bijection between the two sets and this yields the result. Maybe there are a few things to add in between ?
$endgroup$
– Marine Galantin
Jan 15 at 19:02
$begingroup$
in fact that's what I want to do, but I'm not sure if this is true : in fact if I write an attempt for this exercice, I would just have written that there is a bijection between the two sets and this yields the result. Maybe there are a few things to add in between ?
$endgroup$
– Marine Galantin
Jan 15 at 19:02
|
show 4 more comments
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$begingroup$
@Surb No, that is not true.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:15
$begingroup$
@Surb Right, but subgroups of the same order can have different index in general if the group is not finite.
$endgroup$
– Tobias Kildetoft
Jan 15 at 18:19
$begingroup$
So any idea for this exercice ?
$endgroup$
– Marine Galantin
Jan 15 at 18:51
$begingroup$
Can't you just use the automorphism to establish a bijection between the cosets?
$endgroup$
– verret
Jan 15 at 18:55
$begingroup$
in fact that's what I want to do, but I'm not sure if this is true : in fact if I write an attempt for this exercice, I would just have written that there is a bijection between the two sets and this yields the result. Maybe there are a few things to add in between ?
$endgroup$
– Marine Galantin
Jan 15 at 19:02