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The order of conjugate subgroups.
The 2019 Stack Overflow Developer Survey Results Are InDoes the order, lattice of subgroups, and lattice of factor groups, uniquely determine a group up to isomorphism?Question involving Sylow theorems and characteristic subgroupsare all finite index subgroups automorphic to each other?Characteristic subgroups of order $2$how are group order and element order related?Order of an AutomorphismCharacteristic subgroups of normal subgroups are normalExplicit expressions of inner / outer automorphism of symplectic groupAny group automorphism can preserve some nontrivial subgroups if the group is non-abelianFind the order of $operatornameAut(G)$.
$begingroup$
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Then, for any fixed $gin G$, i know that $Hcong gHg^-1$ by inner-automorphism.
I have some question about the 'order':
(1) Is it true that $|H|=|gHg^-1|$ if $|H|<infty$ or $|G:H|<infty$?
(2) If (1) is true, give some counterexample for the case that $|H|=infty$ or $|G:H|=infty$.
Thank you!
group-theory automorphism-group
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Then, for any fixed $gin G$, i know that $Hcong gHg^-1$ by inner-automorphism.
I have some question about the 'order':
(1) Is it true that $|H|=|gHg^-1|$ if $|H|<infty$ or $|G:H|<infty$?
(2) If (1) is true, give some counterexample for the case that $|H|=infty$ or $|G:H|=infty$.
Thank you!
group-theory automorphism-group
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
$endgroup$
1
$begingroup$
It’s true that $|H|=|gHg^-1|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality.
$endgroup$
– Arturo Magidin
Apr 7 at 17:56
add a comment |
$begingroup$
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Then, for any fixed $gin G$, i know that $Hcong gHg^-1$ by inner-automorphism.
I have some question about the 'order':
(1) Is it true that $|H|=|gHg^-1|$ if $|H|<infty$ or $|G:H|<infty$?
(2) If (1) is true, give some counterexample for the case that $|H|=infty$ or $|G:H|=infty$.
Thank you!
group-theory automorphism-group
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
$endgroup$
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Then, for any fixed $gin G$, i know that $Hcong gHg^-1$ by inner-automorphism.
I have some question about the 'order':
(1) Is it true that $|H|=|gHg^-1|$ if $|H|<infty$ or $|G:H|<infty$?
(2) If (1) is true, give some counterexample for the case that $|H|=infty$ or $|G:H|=infty$.
Thank you!
group-theory automorphism-group
group-theory automorphism-group
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
asked Apr 7 at 17:53
PrimaveraPrimavera
32619
32619
1
$begingroup$
It’s true that $|H|=|gHg^-1|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality.
$endgroup$
– Arturo Magidin
Apr 7 at 17:56
add a comment |
1
$begingroup$
It’s true that $|H|=|gHg^-1|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality.
$endgroup$
– Arturo Magidin
Apr 7 at 17:56
1
1
$begingroup$
It’s true that $|H|=|gHg^-1|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality.
$endgroup$
– Arturo Magidin
Apr 7 at 17:56
$begingroup$
It’s true that $|H|=|gHg^-1|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality.
$endgroup$
– Arturo Magidin
Apr 7 at 17:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are three facts needed, here:
- Every group isomorphism is a bijection.
- Given two sets $A$ and $B,$ $A$ and $B$ are of equal cardinality if and only if there exists a bijection between them.
- The order of a group is the cardinality of the set of its elements.
Consequently, we will always have $|H|=left|gHg^-1right|.$
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
$endgroup$
add a comment |
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$begingroup$
There are three facts needed, here:
- Every group isomorphism is a bijection.
- Given two sets $A$ and $B,$ $A$ and $B$ are of equal cardinality if and only if there exists a bijection between them.
- The order of a group is the cardinality of the set of its elements.
Consequently, we will always have $|H|=left|gHg^-1right|.$
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
$endgroup$
add a comment |
$begingroup$
There are three facts needed, here:
- Every group isomorphism is a bijection.
- Given two sets $A$ and $B,$ $A$ and $B$ are of equal cardinality if and only if there exists a bijection between them.
- The order of a group is the cardinality of the set of its elements.
Consequently, we will always have $|H|=left|gHg^-1right|.$
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
$endgroup$
add a comment |
$begingroup$
There are three facts needed, here:
- Every group isomorphism is a bijection.
- Given two sets $A$ and $B,$ $A$ and $B$ are of equal cardinality if and only if there exists a bijection between them.
- The order of a group is the cardinality of the set of its elements.
Consequently, we will always have $|H|=left|gHg^-1right|.$
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
$endgroup$
There are three facts needed, here:
- Every group isomorphism is a bijection.
- Given two sets $A$ and $B,$ $A$ and $B$ are of equal cardinality if and only if there exists a bijection between them.
- The order of a group is the cardinality of the set of its elements.
Consequently, we will always have $|H|=left|gHg^-1right|.$
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
answered Apr 7 at 18:13
Cameron BuieCameron Buie
86.7k773161
86.7k773161
add a comment |
add a comment |
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$begingroup$
It’s true that $|H|=|gHg^-1|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality.
$endgroup$
– Arturo Magidin
Apr 7 at 17:56