One of the conjugacy classes of A4
When I was computing the conjugacy classes of $A_4$, I found that ${(12)(34), (13)(24),(14)(23)}$ is one of the conjugacy classes of $A_4$. This is not hard to see as the center of $A_4$ is trivial. However, we have that
$(23) * (12)(34) * (23) = (13)(24)$
How is this possible? How $(12)(34)$ and $(13)(24)$ can be conjugates if $(23) not in A_4$?
abstract-algebra group-theory permutations symmetric-groups
asked Dec 11 '18 at 6:03
Peter
864
add a comment |
When I was computing the conjugacy classes of $A_4$, I found that ${(12)(34), (13)(24),(14)(23)}$ is one of the conjugacy classes of $A_4$. This is not hard to see as the center of $A_4$ is trivial. However, we have that
$(23) * (12)(34) * (23) = (13)(24)$
How is this possible? How $(12)(34)$ and $(13)(24)$ can be conjugates if $(23) not in A_4$?
abstract-algebra group-theory permutations symmetric-groups
asked Dec 11 '18 at 6:03
Peter
864
add a comment |
When I was computing the conjugacy classes of $A_4$, I found that ${(12)(34), (13)(24),(14)(23)}$ is one of the conjugacy classes of $A_4$. This is not hard to see as the center of $A_4$ is trivial. However, we have that
$(23) * (12)(34) * (23) = (13)(24)$
How is this possible? How $(12)(34)$ and $(13)(24)$ can be conjugates if $(23) not in A_4$?
abstract-algebra group-theory permutations symmetric-groups
asked Dec 11 '18 at 6:03
Peter
864
When I was computing the conjugacy classes of $A_4$, I found that ${(12)(34), (13)(24),(14)(23)}$ is one of the conjugacy classes of $A_4$. This is not hard to see as the center of $A_4$ is trivial. However, we have that
$(23) * (12)(34) * (23) = (13)(24)$
How is this possible? How $(12)(34)$ and $(13)(24)$ can be conjugates if $(23) not in A_4$?
abstract-algebra group-theory permutations symmetric-groups
abstract-algebra group-theory permutations symmetric-groups
asked Dec 11 '18 at 6:03
Peter
864
asked Dec 11 '18 at 6:03
Peter
864
asked Dec 11 '18 at 6:03
Peter
864
asked Dec 11 '18 at 6:03
Peter
864
asked Dec 11 '18 at 6:03
Peter
864
864
add a comment |
add a comment |
1 Answer
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It is possible as the following conjugation relation also holds
$$(132)circ (12)(34)circ (123)=(13)(24)$$
In other words, your two Klein-type (I am referring to the Klein group) elements are conjugates within $A_4$, so there is nothing to worry about. In general, if $G$ is an arbitrary group, $H leqslant G$ a subgroup and $a, b in H$ are conjugates in $H$ that doesn't preclude the possibility of $a$ being conjugated into $b$ by an element in $Gsetminus H$.
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
add a comment |
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It is possible as the following conjugation relation also holds
$$(132)circ (12)(34)circ (123)=(13)(24)$$
In other words, your two Klein-type (I am referring to the Klein group) elements are conjugates within $A_4$, so there is nothing to worry about. In general, if $G$ is an arbitrary group, $H leqslant G$ a subgroup and $a, b in H$ are conjugates in $H$ that doesn't preclude the possibility of $a$ being conjugated into $b$ by an element in $Gsetminus H$.
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
add a comment |
It is possible as the following conjugation relation also holds
$$(132)circ (12)(34)circ (123)=(13)(24)$$
In other words, your two Klein-type (I am referring to the Klein group) elements are conjugates within $A_4$, so there is nothing to worry about. In general, if $G$ is an arbitrary group, $H leqslant G$ a subgroup and $a, b in H$ are conjugates in $H$ that doesn't preclude the possibility of $a$ being conjugated into $b$ by an element in $Gsetminus H$.
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
add a comment |
It is possible as the following conjugation relation also holds
$$(132)circ (12)(34)circ (123)=(13)(24)$$
In other words, your two Klein-type (I am referring to the Klein group) elements are conjugates within $A_4$, so there is nothing to worry about. In general, if $G$ is an arbitrary group, $H leqslant G$ a subgroup and $a, b in H$ are conjugates in $H$ that doesn't preclude the possibility of $a$ being conjugated into $b$ by an element in $Gsetminus H$.
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
It is possible as the following conjugation relation also holds
$$(132)circ (12)(34)circ (123)=(13)(24)$$
In other words, your two Klein-type (I am referring to the Klein group) elements are conjugates within $A_4$, so there is nothing to worry about. In general, if $G$ is an arbitrary group, $H leqslant G$ a subgroup and $a, b in H$ are conjugates in $H$ that doesn't preclude the possibility of $a$ being conjugated into $b$ by an element in $Gsetminus H$.
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
answered Dec 11 '18 at 6:22
ΑΘΩ
2363
2363
add a comment |
add a comment |
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