Mixing
Find all homomorphism from $S_4$ to $z_m$.
0
$begingroup$
Find all homomorphism from $S_4$ to $z_m$.
The only normal subgroup of $S_4$ are trivial , entire group, $A_4$ and klein 4 group.
If m is odd , then there is only one homomorphism (trivial)... What if n is even?
My attempt.
If $6|n$ then there is 3 homomorphism of whose kernal is {e},$A_4$, klein 4 group.
If $6 $ doesn't divide $n$ then there is only 2 homomorphism... Is this correct?
abstract-algebra group-theory group-homomorphism
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
$endgroup$
|
show 1 more comment
0
$begingroup$
Find all homomorphism from $S_4$ to $z_m$.
The only normal subgroup of $S_4$ are trivial , entire group, $A_4$ and klein 4 group.
If m is odd , then there is only one homomorphism (trivial)... What if n is even?
My attempt.
If $6|n$ then there is 3 homomorphism of whose kernal is {e},$A_4$, klein 4 group.
If $6 $ doesn't divide $n$ then there is only 2 homomorphism... Is this correct?
abstract-algebra group-theory group-homomorphism
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
$endgroup$
$begingroup$
Does there exists a homorphism who kernal is klein 4 group?
$endgroup$
– Cloud JR
Jan 23 at 20:49
$begingroup$
Use${ e}$for ${ e}$.
$endgroup$
– Shaun
Jan 23 at 21:03
$begingroup$
For $m=2$ see this question.
$endgroup$
– Dietrich Burde
Jan 23 at 21:14
$begingroup$
Do you know about the abelianization of a group, and in particular what the abelianization of $S_4$ is?
$endgroup$
– Daniel Schepler
Jan 23 at 21:23
1
$begingroup$
OK, then, if I expand the argument I was thinking of, the key points are: 1. Any element of $S_4$ can be written as a product of transpositions. 2. Any two transpositions in $S_4$ are conjugate elements, which must therefore map to the same element $x in mathbb{Z} / langle n rangle$ since $mathbb{Z} / langle n rangle$ is abelian. 3. Since the square of a transposition is the identity, the element $x$ must satisfy $x + x = 0$ (again, in the group $mathbb{Z} / langle n rangle$).
$endgroup$
– Daniel Schepler
Jan 23 at 21:53
|
show 1 more comment
0
0
0
$begingroup$
Find all homomorphism from $S_4$ to $z_m$.
The only normal subgroup of $S_4$ are trivial , entire group, $A_4$ and klein 4 group.
If m is odd , then there is only one homomorphism (trivial)... What if n is even?
My attempt.
If $6|n$ then there is 3 homomorphism of whose kernal is {e},$A_4$, klein 4 group.
If $6 $ doesn't divide $n$ then there is only 2 homomorphism... Is this correct?
abstract-algebra group-theory group-homomorphism
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
$endgroup$
Find all homomorphism from $S_4$ to $z_m$.
The only normal subgroup of $S_4$ are trivial , entire group, $A_4$ and klein 4 group.
If m is odd , then there is only one homomorphism (trivial)... What if n is even?
My attempt.
If $6|n$ then there is 3 homomorphism of whose kernal is {e},$A_4$, klein 4 group.
If $6 $ doesn't divide $n$ then there is only 2 homomorphism... Is this correct?
abstract-algebra group-theory group-homomorphism
abstract-algebra group-theory group-homomorphism
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
asked Jan 23 at 20:45
Cloud JRCloud JR
915518
915518
$begingroup$
Does there exists a homorphism who kernal is klein 4 group?
$endgroup$
– Cloud JR
Jan 23 at 20:49
$begingroup$
Use${ e}$for ${ e}$.
$endgroup$
– Shaun
Jan 23 at 21:03
$begingroup$
For $m=2$ see this question.
$endgroup$
– Dietrich Burde
Jan 23 at 21:14
$begingroup$
Do you know about the abelianization of a group, and in particular what the abelianization of $S_4$ is?
$endgroup$
– Daniel Schepler
Jan 23 at 21:23
1
$begingroup$
OK, then, if I expand the argument I was thinking of, the key points are: 1. Any element of $S_4$ can be written as a product of transpositions. 2. Any two transpositions in $S_4$ are conjugate elements, which must therefore map to the same element $x in mathbb{Z} / langle n rangle$ since $mathbb{Z} / langle n rangle$ is abelian. 3. Since the square of a transposition is the identity, the element $x$ must satisfy $x + x = 0$ (again, in the group $mathbb{Z} / langle n rangle$).
$endgroup$
– Daniel Schepler
Jan 23 at 21:53
|
show 1 more comment
$begingroup$
Does there exists a homorphism who kernal is klein 4 group?
$endgroup$
– Cloud JR
Jan 23 at 20:49
$begingroup$
Use${ e}$for ${ e}$.
$endgroup$
– Shaun
Jan 23 at 21:03
$begingroup$
For $m=2$ see this question.
$endgroup$
– Dietrich Burde
Jan 23 at 21:14
$begingroup$
Do you know about the abelianization of a group, and in particular what the abelianization of $S_4$ is?
$endgroup$
– Daniel Schepler
Jan 23 at 21:23
1
$begingroup$
OK, then, if I expand the argument I was thinking of, the key points are: 1. Any element of $S_4$ can be written as a product of transpositions. 2. Any two transpositions in $S_4$ are conjugate elements, which must therefore map to the same element $x in mathbb{Z} / langle n rangle$ since $mathbb{Z} / langle n rangle$ is abelian. 3. Since the square of a transposition is the identity, the element $x$ must satisfy $x + x = 0$ (again, in the group $mathbb{Z} / langle n rangle$).
$endgroup$
– Daniel Schepler
Jan 23 at 21:53
$begingroup$
Does there exists a homorphism who kernal is klein 4 group?
$endgroup$
– Cloud JR
Jan 23 at 20:49
$begingroup$
Does there exists a homorphism who kernal is klein 4 group?
$endgroup$
– Cloud JR
Jan 23 at 20:49
$begingroup$
Use
${ e}$ for ${ e}$.$endgroup$
– Shaun
Jan 23 at 21:03
$begingroup$
Use
${ e}$ for ${ e}$.$endgroup$
– Shaun
Jan 23 at 21:03
$begingroup$
For $m=2$ see this question.
$endgroup$
– Dietrich Burde
Jan 23 at 21:14
$begingroup$
For $m=2$ see this question.
$endgroup$
– Dietrich Burde
Jan 23 at 21:14
$begingroup$
Do you know about the abelianization of a group, and in particular what the abelianization of $S_4$ is?
$endgroup$
– Daniel Schepler
Jan 23 at 21:23
$begingroup$
Do you know about the abelianization of a group, and in particular what the abelianization of $S_4$ is?
$endgroup$
– Daniel Schepler
Jan 23 at 21:23
1
1
$begingroup$
OK, then, if I expand the argument I was thinking of, the key points are: 1. Any element of $S_4$ can be written as a product of transpositions. 2. Any two transpositions in $S_4$ are conjugate elements, which must therefore map to the same element $x in mathbb{Z} / langle n rangle$ since $mathbb{Z} / langle n rangle$ is abelian. 3. Since the square of a transposition is the identity, the element $x$ must satisfy $x + x = 0$ (again, in the group $mathbb{Z} / langle n rangle$).
$endgroup$
– Daniel Schepler
Jan 23 at 21:53
$begingroup$
OK, then, if I expand the argument I was thinking of, the key points are: 1. Any element of $S_4$ can be written as a product of transpositions. 2. Any two transpositions in $S_4$ are conjugate elements, which must therefore map to the same element $x in mathbb{Z} / langle n rangle$ since $mathbb{Z} / langle n rangle$ is abelian. 3. Since the square of a transposition is the identity, the element $x$ must satisfy $x + x = 0$ (again, in the group $mathbb{Z} / langle n rangle$).
$endgroup$
– Daniel Schepler
Jan 23 at 21:53
|
show 1 more comment
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$begingroup$
Does there exists a homorphism who kernal is klein 4 group?
$endgroup$
– Cloud JR
Jan 23 at 20:49
$begingroup$
Use
${ e}$for ${ e}$.$endgroup$
– Shaun
Jan 23 at 21:03
$begingroup$
For $m=2$ see this question.
$endgroup$
– Dietrich Burde
Jan 23 at 21:14
$begingroup$
Do you know about the abelianization of a group, and in particular what the abelianization of $S_4$ is?
$endgroup$
– Daniel Schepler
Jan 23 at 21:23
1
$begingroup$
OK, then, if I expand the argument I was thinking of, the key points are: 1. Any element of $S_4$ can be written as a product of transpositions. 2. Any two transpositions in $S_4$ are conjugate elements, which must therefore map to the same element $x in mathbb{Z} / langle n rangle$ since $mathbb{Z} / langle n rangle$ is abelian. 3. Since the square of a transposition is the identity, the element $x$ must satisfy $x + x = 0$ (again, in the group $mathbb{Z} / langle n rangle$).
$endgroup$
– Daniel Schepler
Jan 23 at 21:53