Construct a non trivial homomorphism from $mathbb{Z}_2timesmathbb{Z}_4$ to $mathbb{Z}_8$
I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:
$(0,0)$: $1$
$(0,1)$: $4$
$(1,0)$: $2$
$(1,1)$: $4$
$(0,2)$: $2$
$(1,2)$: $2$
$(0,3)$: $4$
$(1,3)$: $4$
Can anyone help me?
abstract-algebra group-theory cyclic-groups group-homomorphism direct-product
edited Jan 4 at 18:16
Shaun
8,820113681
asked Jan 4 at 17:23
FailousaFailousa
31
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Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:
$(0,0)$: $1$
$(0,1)$: $4$
$(1,0)$: $2$
$(1,1)$: $4$
$(0,2)$: $2$
$(1,2)$: $2$
$(0,3)$: $4$
$(1,3)$: $4$
Can anyone help me?
abstract-algebra group-theory cyclic-groups group-homomorphism direct-product
edited Jan 4 at 18:16
Shaun
8,820113681
asked Jan 4 at 17:23
FailousaFailousa
31
New contributor
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:
$(0,0)$: $1$
$(0,1)$: $4$
$(1,0)$: $2$
$(1,1)$: $4$
$(0,2)$: $2$
$(1,2)$: $2$
$(0,3)$: $4$
$(1,3)$: $4$
Can anyone help me?
abstract-algebra group-theory cyclic-groups group-homomorphism direct-product
edited Jan 4 at 18:16
Shaun
8,820113681
asked Jan 4 at 17:23
FailousaFailousa
31
New contributor
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I don't know how to solve this problem since the group $mathbb{Z}_2timesmathbb{Z}_4$ is not cyclic. I just know that $mathbb{Z}_2 times mathbb{Z}_4= { (0,0), (0,1),(1,0),(1,1),(0,2),(1,2),(0,3),(1,3) }$ and that the orders of these elements are:
$(0,0)$: $1$
$(0,1)$: $4$
$(1,0)$: $2$
$(1,1)$: $4$
$(0,2)$: $2$
$(1,2)$: $2$
$(0,3)$: $4$
$(1,3)$: $4$
Can anyone help me?
abstract-algebra group-theory cyclic-groups group-homomorphism direct-product
abstract-algebra group-theory cyclic-groups group-homomorphism direct-product
edited Jan 4 at 18:16
Shaun
8,820113681
asked Jan 4 at 17:23
FailousaFailousa
31
New contributor
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 18:16
Shaun
8,820113681
asked Jan 4 at 17:23
FailousaFailousa
31
New contributor
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 18:16
Shaun
8,820113681
edited Jan 4 at 18:16
Shaun
8,820113681
edited Jan 4 at 18:16
Shaun
8,820113681
8,820113681
asked Jan 4 at 17:23
FailousaFailousa
31
New contributor
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 17:23
FailousaFailousa
31
asked Jan 4 at 17:23
FailousaFailousa
31
31
New contributor
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Failousa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
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A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.
answered Jan 4 at 17:26
user3482749user3482749
2,979414
Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
– Failousa
Jan 4 at 19:43
$f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
– user3482749
Jan 4 at 19:46
Ok.Thank you very much!
– Failousa
Jan 4 at 19:53
add a comment |
Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
$$
(x,y)mapsto alpha(x)+beta(y)
$$
where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.
Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?
answered Jan 4 at 17:38
egregegreg
179k1485202
I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
– Failousa
Jan 4 at 19:58
@Failousa You can prove it directly.
– egreg
Jan 4 at 20:25
add a comment |
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2 Answers
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2 Answers
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A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.
answered Jan 4 at 17:26
user3482749user3482749
2,979414
Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
– Failousa
Jan 4 at 19:43
$f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
– user3482749
Jan 4 at 19:46
Ok.Thank you very much!
– Failousa
Jan 4 at 19:53
add a comment |
A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.
answered Jan 4 at 17:26
user3482749user3482749
2,979414
Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
– Failousa
Jan 4 at 19:43
$f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
– user3482749
Jan 4 at 19:46
Ok.Thank you very much!
– Failousa
Jan 4 at 19:53
add a comment |
A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.
answered Jan 4 at 17:26
user3482749user3482749
2,979414
A homomorphism out of $mathbb{Z}_2 times mathbb{Z}_4$ is determined by where it sends $(1,0)$ and $(0,1)$. To make it non-trivial, you just need to send one of those to somewhere that isn't $(0,0)$. Note that you must send $(1,0)$ to an element of order $1$ or $2$, and $(0,1)$ to an element of order $1$, $2$, or $4$, or the resulting function will not be a homomorphism.
answered Jan 4 at 17:26
user3482749user3482749
2,979414
answered Jan 4 at 17:26
user3482749user3482749
2,979414
answered Jan 4 at 17:26
user3482749user3482749
2,979414
answered Jan 4 at 17:26
user3482749user3482749
2,979414
2,979414
Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
– Failousa
Jan 4 at 19:43
$f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
– user3482749
Jan 4 at 19:46
Ok.Thank you very much!
– Failousa
Jan 4 at 19:53
add a comment |
Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
– Failousa
Jan 4 at 19:43
$f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
– user3482749
Jan 4 at 19:46
Ok.Thank you very much!
– Failousa
Jan 4 at 19:53
Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
– Failousa
Jan 4 at 19:43
Thank you for your help.So,I just have to send (1,0) to 0 or 4 and (0,1) to 0 or 4 or 2 or 6 but without both being 0 at the same time.Right? If I choose for example f(1,0)=0 and f(0,1)=2 I want to show that f(0,0)=0$in mathbb{Z_8}$ in order f to be an homomorphism.So, f(0,0)=f(2,4)=2f(1,2).We know that (1,2)=(1,0)+2(0,1).How can I compute f(1,2)?Is it correct to say that f(1,2)=0+2$cdot$2=4 ,so f(0,0)=2$cdot$4=8=0 in $mathbb{Z_8}$? I don't know if my thoughts are right.
– Failousa
Jan 4 at 19:43
$f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
– user3482749
Jan 4 at 19:46
$f(1,2) = f(1,0) + 2f(0,1)$, since $f$ is a homomorphism, so $f(1,2) = 2(2) = 4$. You don't need to prove that $f(0,0) = 0$: you defined it to be that.
– user3482749
Jan 4 at 19:46
Ok.Thank you very much!
– Failousa
Jan 4 at 19:53
Ok.Thank you very much!
– Failousa
Jan 4 at 19:53
add a comment |
Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
$$
(x,y)mapsto alpha(x)+beta(y)
$$
where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.
Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?
answered Jan 4 at 17:38
egregegreg
179k1485202
I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
– Failousa
Jan 4 at 19:58
@Failousa You can prove it directly.
– egreg
Jan 4 at 20:25
add a comment |
Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
$$
(x,y)mapsto alpha(x)+beta(y)
$$
where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.
Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?
answered Jan 4 at 17:38
egregegreg
179k1485202
I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
– Failousa
Jan 4 at 19:58
@Failousa You can prove it directly.
– egreg
Jan 4 at 20:25
add a comment |
Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
$$
(x,y)mapsto alpha(x)+beta(y)
$$
where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.
Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?
answered Jan 4 at 17:38
egregegreg
179k1485202
Hint: If $G_1$, $G_2$ and $G_3$ are additive abelian groups, a homomorphism $G_1times G_2to G_3$ is of the form
$$
(x,y)mapsto alpha(x)+beta(y)
$$
where $alphacolon G_1to G_3$ and $betacolon G_2to G_3$ are (arbitrary) homomorphisms. This is not trivial so long as either $alpha$ or $beta$ is not trivial.
Can you find a nontrivial homomorphism $mathbb{Z}_2tomathbb{Z}_8$?
answered Jan 4 at 17:38
egregegreg
179k1485202
answered Jan 4 at 17:38
egregegreg
179k1485202
answered Jan 4 at 17:38
egregegreg
179k1485202
answered Jan 4 at 17:38
egregegreg
179k1485202
179k1485202
I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
– Failousa
Jan 4 at 19:58
@Failousa You can prove it directly.
– egreg
Jan 4 at 20:25
add a comment |
I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
– Failousa
Jan 4 at 19:58
@Failousa You can prove it directly.
– egreg
Jan 4 at 20:25
I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
– Failousa
Jan 4 at 19:58
I didn't know that.Thanks for your help!Yes,I can find a nontrivial homomorphism $mathbb{Z_2}tomathbb{Z_8}$ but can I have a proof of what you wrote? Or a link where I can find it?
– Failousa
Jan 4 at 19:58
@Failousa You can prove it directly.
– egreg
Jan 4 at 20:25
@Failousa You can prove it directly.
– egreg
Jan 4 at 20:25
add a comment |
Failousa is a new contributor. Be nice, and check out our Code of Conduct.
Failousa is a new contributor. Be nice, and check out our Code of Conduct.
Failousa is a new contributor. Be nice, and check out our Code of Conduct.
Failousa is a new contributor. Be nice, and check out our Code of Conduct.
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