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 : $$[G: Psi(H) ] = [G:H] $$











share|cite|improve this question











$endgroup$



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
















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 : $$[G: Psi(H) ] = [G:H] $$











share|cite|improve this question











$endgroup$



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














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 : $$[G: Psi(H) ] = [G:H] $$











share|cite|improve this question











$endgroup$




$G$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 18:08









Surb

38.4k94478




38.4k94478










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


















  • $begingroup$
    @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










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