Mixing
How many homomorphisms from Z4 to S4.
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Let $Æ: mathbbZ_4to S_4$ be a random homomorphism.
From theory I have:
- Since $1_mathbbZ_4$ is the identity element in $mathbbZ_4$, $Æ(1_mathbbZ_4)=1_s_4$.
- $Æ(1^-1)=Æ(3)=(Æ(1))^-1, Æ(2^-1)=Æ(2)=(Æ(2))^-1$
These don't seem enough to find all homomorphisms. What am i missing?
abstract-algebra group-homomorphism
asked Jul 20 at 11:49
Sovengarde
112
add a comment |Â
up vote
1
down vote
favorite
Let $Æ: mathbbZ_4to S_4$ be a random homomorphism.
From theory I have:
- Since $1_mathbbZ_4$ is the identity element in $mathbbZ_4$, $Æ(1_mathbbZ_4)=1_s_4$.
- $Æ(1^-1)=Æ(3)=(Æ(1))^-1, Æ(2^-1)=Æ(2)=(Æ(2))^-1$
These don't seem enough to find all homomorphisms. What am i missing?
abstract-algebra group-homomorphism
asked Jul 20 at 11:49
Sovengarde
112
I'm a bit confused by the notation. Is $1_mathbb Z_4$ what is usually called $0$? A hint on finding the homomorphisms: 1) find all possible quotient groups of $mathbb Z_4$ (there are 3 of them), and 2) find all subgroups of $S_4$ isomorphic to some quotient of $mathbb Z_4$.
– lisyarus
Jul 20 at 11:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $Æ: mathbbZ_4to S_4$ be a random homomorphism.
From theory I have:
- Since $1_mathbbZ_4$ is the identity element in $mathbbZ_4$, $Æ(1_mathbbZ_4)=1_s_4$.
- $Æ(1^-1)=Æ(3)=(Æ(1))^-1, Æ(2^-1)=Æ(2)=(Æ(2))^-1$
These don't seem enough to find all homomorphisms. What am i missing?
abstract-algebra group-homomorphism
asked Jul 20 at 11:49
Sovengarde
112
Let $Æ: mathbbZ_4to S_4$ be a random homomorphism.
From theory I have:
- Since $1_mathbbZ_4$ is the identity element in $mathbbZ_4$, $Æ(1_mathbbZ_4)=1_s_4$.
- $Æ(1^-1)=Æ(3)=(Æ(1))^-1, Æ(2^-1)=Æ(2)=(Æ(2))^-1$
These don't seem enough to find all homomorphisms. What am i missing?
abstract-algebra group-homomorphism
asked Jul 20 at 11:49
Sovengarde
112
asked Jul 20 at 11:49
Sovengarde
112
asked Jul 20 at 11:49
Sovengarde
112
asked Jul 20 at 11:49
Sovengarde
112
112
I'm a bit confused by the notation. Is $1_mathbb Z_4$ what is usually called $0$? A hint on finding the homomorphisms: 1) find all possible quotient groups of $mathbb Z_4$ (there are 3 of them), and 2) find all subgroups of $S_4$ isomorphic to some quotient of $mathbb Z_4$.
– lisyarus
Jul 20 at 11:54
add a comment |Â
I'm a bit confused by the notation. Is $1_mathbb Z_4$ what is usually called $0$? A hint on finding the homomorphisms: 1) find all possible quotient groups of $mathbb Z_4$ (there are 3 of them), and 2) find all subgroups of $S_4$ isomorphic to some quotient of $mathbb Z_4$.
– lisyarus
Jul 20 at 11:54
I'm a bit confused by the notation. Is $1_mathbb Z_4$ what is usually called $0$? A hint on finding the homomorphisms: 1) find all possible quotient groups of $mathbb Z_4$ (there are 3 of them), and 2) find all subgroups of $S_4$ isomorphic to some quotient of $mathbb Z_4$.
– lisyarus
Jul 20 at 11:54
I'm a bit confused by the notation. Is $1_mathbb Z_4$ what is usually called $0$? A hint on finding the homomorphisms: 1) find all possible quotient groups of $mathbb Z_4$ (there are 3 of them), and 2) find all subgroups of $S_4$ isomorphic to some quotient of $mathbb Z_4$.
– lisyarus
Jul 20 at 11:54
add a comment |Â
1 Answer
1
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1
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There are problems with your calculation. You seem to be confusing the additive and multiplicative structure on $mathbbZ_4$. If you say that the inverse of 1 is 3, then this means you consider the additive structure. But then the unit element is 0, not 1.
But I can answer your question. Let $mathbbZ_4$ be the ADDITIVE group of residue classes $pmod 4$. Then the image of $1$ uniquely determines a homomorphism from $mathbbZ_4$ to another group. This image can be any element whose order divides 4, and exactly those. So your question can be rephrased as follows: how many elements are there in $S_4$ whose order is 1, 2 or 4?
There is exactly 1 element of order 1, the identity.
There are $binom42= 6$ transpositions and 3 permutations that are products of two disjoint transpositions. So there are 9 elements of order 2.
Finally, there are 6 elements of order 4 in $S_4$.
So the answer is: there are $1+9+6= 16$ elements of order 1, 2 or 4 in $S_4$, hence 16 homomorphisms from $mathbbZ_4$ into $S_4$.
answered Jul 20 at 12:00
A. Pongrácz
2,339221
"Then the image of 1 uniquely determines a homomorphism from Z4 to another group. This image can be any element whose order divides 4, and exactly those." Can you elaborate a bit more?
– Sovengarde
Jul 20 at 12:12
The elements of $mathbbZ_4$ are $1, 1+1, 1+1+1, 1+1+1+1$. Applying $varphi$ and using the property that it is a homomorphism yields unicity once $varphi(1)$ is fixed. Think about the rest!
– A. Pongrácz
Jul 20 at 12:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There are problems with your calculation. You seem to be confusing the additive and multiplicative structure on $mathbbZ_4$. If you say that the inverse of 1 is 3, then this means you consider the additive structure. But then the unit element is 0, not 1.
But I can answer your question. Let $mathbbZ_4$ be the ADDITIVE group of residue classes $pmod 4$. Then the image of $1$ uniquely determines a homomorphism from $mathbbZ_4$ to another group. This image can be any element whose order divides 4, and exactly those. So your question can be rephrased as follows: how many elements are there in $S_4$ whose order is 1, 2 or 4?
There is exactly 1 element of order 1, the identity.
There are $binom42= 6$ transpositions and 3 permutations that are products of two disjoint transpositions. So there are 9 elements of order 2.
Finally, there are 6 elements of order 4 in $S_4$.
So the answer is: there are $1+9+6= 16$ elements of order 1, 2 or 4 in $S_4$, hence 16 homomorphisms from $mathbbZ_4$ into $S_4$.
answered Jul 20 at 12:00
A. Pongrácz
2,339221
"Then the image of 1 uniquely determines a homomorphism from Z4 to another group. This image can be any element whose order divides 4, and exactly those." Can you elaborate a bit more?
– Sovengarde
Jul 20 at 12:12
The elements of $mathbbZ_4$ are $1, 1+1, 1+1+1, 1+1+1+1$. Applying $varphi$ and using the property that it is a homomorphism yields unicity once $varphi(1)$ is fixed. Think about the rest!
– A. Pongrácz
Jul 20 at 12:15
add a comment |Â
up vote
1
down vote
accepted
There are problems with your calculation. You seem to be confusing the additive and multiplicative structure on $mathbbZ_4$. If you say that the inverse of 1 is 3, then this means you consider the additive structure. But then the unit element is 0, not 1.
But I can answer your question. Let $mathbbZ_4$ be the ADDITIVE group of residue classes $pmod 4$. Then the image of $1$ uniquely determines a homomorphism from $mathbbZ_4$ to another group. This image can be any element whose order divides 4, and exactly those. So your question can be rephrased as follows: how many elements are there in $S_4$ whose order is 1, 2 or 4?
There is exactly 1 element of order 1, the identity.
There are $binom42= 6$ transpositions and 3 permutations that are products of two disjoint transpositions. So there are 9 elements of order 2.
Finally, there are 6 elements of order 4 in $S_4$.
So the answer is: there are $1+9+6= 16$ elements of order 1, 2 or 4 in $S_4$, hence 16 homomorphisms from $mathbbZ_4$ into $S_4$.
answered Jul 20 at 12:00
A. Pongrácz
2,339221
"Then the image of 1 uniquely determines a homomorphism from Z4 to another group. This image can be any element whose order divides 4, and exactly those." Can you elaborate a bit more?
– Sovengarde
Jul 20 at 12:12
The elements of $mathbbZ_4$ are $1, 1+1, 1+1+1, 1+1+1+1$. Applying $varphi$ and using the property that it is a homomorphism yields unicity once $varphi(1)$ is fixed. Think about the rest!
– A. Pongrácz
Jul 20 at 12:15
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There are problems with your calculation. You seem to be confusing the additive and multiplicative structure on $mathbbZ_4$. If you say that the inverse of 1 is 3, then this means you consider the additive structure. But then the unit element is 0, not 1.
But I can answer your question. Let $mathbbZ_4$ be the ADDITIVE group of residue classes $pmod 4$. Then the image of $1$ uniquely determines a homomorphism from $mathbbZ_4$ to another group. This image can be any element whose order divides 4, and exactly those. So your question can be rephrased as follows: how many elements are there in $S_4$ whose order is 1, 2 or 4?
There is exactly 1 element of order 1, the identity.
There are $binom42= 6$ transpositions and 3 permutations that are products of two disjoint transpositions. So there are 9 elements of order 2.
Finally, there are 6 elements of order 4 in $S_4$.
So the answer is: there are $1+9+6= 16$ elements of order 1, 2 or 4 in $S_4$, hence 16 homomorphisms from $mathbbZ_4$ into $S_4$.
answered Jul 20 at 12:00
A. Pongrácz
2,339221
There are problems with your calculation. You seem to be confusing the additive and multiplicative structure on $mathbbZ_4$. If you say that the inverse of 1 is 3, then this means you consider the additive structure. But then the unit element is 0, not 1.
But I can answer your question. Let $mathbbZ_4$ be the ADDITIVE group of residue classes $pmod 4$. Then the image of $1$ uniquely determines a homomorphism from $mathbbZ_4$ to another group. This image can be any element whose order divides 4, and exactly those. So your question can be rephrased as follows: how many elements are there in $S_4$ whose order is 1, 2 or 4?
There is exactly 1 element of order 1, the identity.
There are $binom42= 6$ transpositions and 3 permutations that are products of two disjoint transpositions. So there are 9 elements of order 2.
Finally, there are 6 elements of order 4 in $S_4$.
So the answer is: there are $1+9+6= 16$ elements of order 1, 2 or 4 in $S_4$, hence 16 homomorphisms from $mathbbZ_4$ into $S_4$.
answered Jul 20 at 12:00
A. Pongrácz
2,339221
answered Jul 20 at 12:00
A. Pongrácz
2,339221
answered Jul 20 at 12:00
A. Pongrácz
2,339221
answered Jul 20 at 12:00
A. Pongrácz
2,339221
2,339221
"Then the image of 1 uniquely determines a homomorphism from Z4 to another group. This image can be any element whose order divides 4, and exactly those." Can you elaborate a bit more?
– Sovengarde
Jul 20 at 12:12
The elements of $mathbbZ_4$ are $1, 1+1, 1+1+1, 1+1+1+1$. Applying $varphi$ and using the property that it is a homomorphism yields unicity once $varphi(1)$ is fixed. Think about the rest!
– A. Pongrácz
Jul 20 at 12:15
add a comment |Â
"Then the image of 1 uniquely determines a homomorphism from Z4 to another group. This image can be any element whose order divides 4, and exactly those." Can you elaborate a bit more?
– Sovengarde
Jul 20 at 12:12
The elements of $mathbbZ_4$ are $1, 1+1, 1+1+1, 1+1+1+1$. Applying $varphi$ and using the property that it is a homomorphism yields unicity once $varphi(1)$ is fixed. Think about the rest!
– A. Pongrácz
Jul 20 at 12:15
"Then the image of 1 uniquely determines a homomorphism from Z4 to another group. This image can be any element whose order divides 4, and exactly those." Can you elaborate a bit more?
– Sovengarde
Jul 20 at 12:12
"Then the image of 1 uniquely determines a homomorphism from Z4 to another group. This image can be any element whose order divides 4, and exactly those." Can you elaborate a bit more?
– Sovengarde
Jul 20 at 12:12
The elements of $mathbbZ_4$ are $1, 1+1, 1+1+1, 1+1+1+1$. Applying $varphi$ and using the property that it is a homomorphism yields unicity once $varphi(1)$ is fixed. Think about the rest!
– A. Pongrácz
Jul 20 at 12:15
The elements of $mathbbZ_4$ are $1, 1+1, 1+1+1, 1+1+1+1$. Applying $varphi$ and using the property that it is a homomorphism yields unicity once $varphi(1)$ is fixed. Think about the rest!
– A. Pongrácz
Jul 20 at 12:15
add a comment |Â
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I'm a bit confused by the notation. Is $1_mathbb Z_4$ what is usually called $0$? A hint on finding the homomorphisms: 1) find all possible quotient groups of $mathbb Z_4$ (there are 3 of them), and 2) find all subgroups of $S_4$ isomorphic to some quotient of $mathbb Z_4$.
– lisyarus
Jul 20 at 11:54