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Are two isomorphic finite subgroups of $SO(4)$ conjugate?

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3

0

$begingroup$

Let $A,B$ be two finite subgroups of $SO(4)$ such that $A$ and $B$ are isomorphic as abstract groups. Can we find a $g in SO(4)$ such that
$$
gAg^-1=B?
$$
If it is the case, does the same conclusion hold for $SO(n)$?

matrices group-theory finite-groups orthogonal-matrices

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asked Mar 21 at 13:47

TotoroTotoro

27415

$endgroup$

add a comment | 

3

0

$begingroup$

Let $A,B$ be two finite subgroups of $SO(4)$ such that $A$ and $B$ are isomorphic as abstract groups. Can we find a $g in SO(4)$ such that
$$
gAg^-1=B?
$$
If it is the case, does the same conclusion hold for $SO(n)$?

matrices group-theory finite-groups orthogonal-matrices

share|cite|improve this question

asked Mar 21 at 13:47

TotoroTotoro

27415

$endgroup$

add a comment | 

$begingroup$

Let $A,B$ be two finite subgroups of $SO(4)$ such that $A$ and $B$ are isomorphic as abstract groups. Can we find a $g in SO(4)$ such that
$$
gAg^-1=B?
$$
If it is the case, does the same conclusion hold for $SO(n)$?

matrices group-theory finite-groups orthogonal-matrices

share|cite|improve this question

asked Mar 21 at 13:47

TotoroTotoro

27415

$endgroup$

share|cite|improve this question

asked Mar 21 at 13:47

TotoroTotoro

27415

share|cite|improve this question

asked Mar 21 at 13:47

TotoroTotoro

27415

asked Mar 21 at 13:47

TotoroTotoro

27415

asked Mar 21 at 13:47

TotoroTotoro

27415

1 Answer
1

active

oldest

votes

4

$begingroup$

$newcommandmat[1]left(beginmatrix#1endmatrixright)$
Identifying $SO(4)$ with the group of deteminant-one matrices $Qin mathrmM_4(mathbb R)$ such that $Q^TQ=QQ^T=mathrmId$, take

$$A=lbracemathrmId,mat-1\&-1\&&-1\&&&-1rbracequadtextandquad
B=lbracemathrmId,mat1\&-1\&&1\&&&-1rbrace.$$

Then both are subgroups of order $2$, and they cannot be conjugate, since their non-trivial elements have different characteristic polynomials.

Edited: The group $A$ was changed following a comment by TastyRomeo, the original group was a subgroup of $O(4)$, rather than of $SO(4)$.

share|cite|improve this answer

edited Mar 21 at 14:16

answered Mar 21 at 14:07

kneidellkneidell

1,7631321

$endgroup$

add a comment | 

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4

$begingroup$

$newcommandmat[1]left(beginmatrix#1endmatrixright)$
Identifying $SO(4)$ with the group of deteminant-one matrices $Qin mathrmM_4(mathbb R)$ such that $Q^TQ=QQ^T=mathrmId$, take

$$A=lbracemathrmId,mat-1\&-1\&&-1\&&&-1rbracequadtextandquad
B=lbracemathrmId,mat1\&-1\&&1\&&&-1rbrace.$$

Then both are subgroups of order $2$, and they cannot be conjugate, since their non-trivial elements have different characteristic polynomials.

Edited: The group $A$ was changed following a comment by TastyRomeo, the original group was a subgroup of $O(4)$, rather than of $SO(4)$.

share|cite|improve this answer

edited Mar 21 at 14:16

answered Mar 21 at 14:07

kneidellkneidell

1,7631321

$endgroup$

add a comment | 

4

$begingroup$

$newcommandmat[1]left(beginmatrix#1endmatrixright)$
Identifying $SO(4)$ with the group of deteminant-one matrices $Qin mathrmM_4(mathbb R)$ such that $Q^TQ=QQ^T=mathrmId$, take

$$A=lbracemathrmId,mat-1\&-1\&&-1\&&&-1rbracequadtextandquad
B=lbracemathrmId,mat1\&-1\&&1\&&&-1rbrace.$$

Then both are subgroups of order $2$, and they cannot be conjugate, since their non-trivial elements have different characteristic polynomials.

Edited: The group $A$ was changed following a comment by TastyRomeo, the original group was a subgroup of $O(4)$, rather than of $SO(4)$.

share|cite|improve this answer

edited Mar 21 at 14:16

answered Mar 21 at 14:07

kneidellkneidell

1,7631321

$endgroup$

add a comment | 

$begingroup$

$newcommandmat[1]left(beginmatrix#1endmatrixright)$
Identifying $SO(4)$ with the group of deteminant-one matrices $Qin mathrmM_4(mathbb R)$ such that $Q^TQ=QQ^T=mathrmId$, take

$$A=lbracemathrmId,mat-1\&-1\&&-1\&&&-1rbracequadtextandquad
B=lbracemathrmId,mat1\&-1\&&1\&&&-1rbrace.$$

Then both are subgroups of order $2$, and they cannot be conjugate, since their non-trivial elements have different characteristic polynomials.

Edited: The group $A$ was changed following a comment by TastyRomeo, the original group was a subgroup of $O(4)$, rather than of $SO(4)$.

share|cite|improve this answer

edited Mar 21 at 14:16

answered Mar 21 at 14:07

kneidellkneidell

1,7631321

$endgroup$

share|cite|improve this answer

edited Mar 21 at 14:16

answered Mar 21 at 14:07

kneidellkneidell

1,7631321

answered Mar 21 at 14:07

kneidellkneidell

1,7631321

answered Mar 21 at 14:07

kneidellkneidell

1,7631321