Mixing
Verifying a statement related to group homomorphism
$begingroup$
I am studying from a book about group-theory. I got the chapter of normal groups and isomorphisms. There was a question:
Let $X=mathbb{Z}_{4}$,$X'={0,2}$,$Y=mathbb{Z}_{2}timesmathbb{Z}_{2}$ and $Y'={(0,0),(1,1)}$.
Verify: $X'cong Y'$ and $X/X'cong Y/Y'$, but $Xnotcong Y$.
Where do I start? But verifying I guess it possible to prove it with one of the isomorphisms laws and I should check it manually. But How should I do it?
group-theory group-homomorphism
asked Jan 7 at 18:15
vesiivesii
936
$endgroup$
add a comment |
$begingroup$
I am studying from a book about group-theory. I got the chapter of normal groups and isomorphisms. There was a question:
Let $X=mathbb{Z}_{4}$,$X'={0,2}$,$Y=mathbb{Z}_{2}timesmathbb{Z}_{2}$ and $Y'={(0,0),(1,1)}$.
Verify: $X'cong Y'$ and $X/X'cong Y/Y'$, but $Xnotcong Y$.
Where do I start? But verifying I guess it possible to prove it with one of the isomorphisms laws and I should check it manually. But How should I do it?
group-theory group-homomorphism
asked Jan 7 at 18:15
vesiivesii
936
$endgroup$
$begingroup$
Is $X^prime$ cyclic? What about $Y^prime$? And how could you describe the isomorphism $X^prime cong Y^prime$? What distinguishes $X$ and $Y$?
$endgroup$
– ÍgjøgnumMeg
Jan 7 at 18:27
$begingroup$
@ÍgjøgnumMeg I think they are, but how does it help us?
$endgroup$
– vesii
Jan 8 at 0:45
$begingroup$
A cyclic group is generated by a single element, so the image of a homomorphism out of a cyclic group is entirely determined by the image of a generator for that group.
$endgroup$
– ÍgjøgnumMeg
Jan 8 at 8:23
$begingroup$
@ÍgjøgnumMeg I still don't see it. can you explain?
$endgroup$
– vesii
Jan 12 at 14:40
add a comment |
$begingroup$
I am studying from a book about group-theory. I got the chapter of normal groups and isomorphisms. There was a question:
Let $X=mathbb{Z}_{4}$,$X'={0,2}$,$Y=mathbb{Z}_{2}timesmathbb{Z}_{2}$ and $Y'={(0,0),(1,1)}$.
Verify: $X'cong Y'$ and $X/X'cong Y/Y'$, but $Xnotcong Y$.
Where do I start? But verifying I guess it possible to prove it with one of the isomorphisms laws and I should check it manually. But How should I do it?
group-theory group-homomorphism
asked Jan 7 at 18:15
vesiivesii
936
$endgroup$
I am studying from a book about group-theory. I got the chapter of normal groups and isomorphisms. There was a question:
Let $X=mathbb{Z}_{4}$,$X'={0,2}$,$Y=mathbb{Z}_{2}timesmathbb{Z}_{2}$ and $Y'={(0,0),(1,1)}$.
Verify: $X'cong Y'$ and $X/X'cong Y/Y'$, but $Xnotcong Y$.
Where do I start? But verifying I guess it possible to prove it with one of the isomorphisms laws and I should check it manually. But How should I do it?
group-theory group-homomorphism
group-theory group-homomorphism
asked Jan 7 at 18:15
vesiivesii
936
asked Jan 7 at 18:15
vesiivesii
936
asked Jan 7 at 18:15
vesiivesii
936
asked Jan 7 at 18:15
vesiivesii
936
asked Jan 7 at 18:15
vesiivesii
936
936
$begingroup$
Is $X^prime$ cyclic? What about $Y^prime$? And how could you describe the isomorphism $X^prime cong Y^prime$? What distinguishes $X$ and $Y$?
$endgroup$
– ÍgjøgnumMeg
Jan 7 at 18:27
$begingroup$
@ÍgjøgnumMeg I think they are, but how does it help us?
$endgroup$
– vesii
Jan 8 at 0:45
$begingroup$
A cyclic group is generated by a single element, so the image of a homomorphism out of a cyclic group is entirely determined by the image of a generator for that group.
$endgroup$
– ÍgjøgnumMeg
Jan 8 at 8:23
$begingroup$
@ÍgjøgnumMeg I still don't see it. can you explain?
$endgroup$
– vesii
Jan 12 at 14:40
add a comment |
$begingroup$
Is $X^prime$ cyclic? What about $Y^prime$? And how could you describe the isomorphism $X^prime cong Y^prime$? What distinguishes $X$ and $Y$?
$endgroup$
– ÍgjøgnumMeg
Jan 7 at 18:27
$begingroup$
@ÍgjøgnumMeg I think they are, but how does it help us?
$endgroup$
– vesii
Jan 8 at 0:45
$begingroup$
A cyclic group is generated by a single element, so the image of a homomorphism out of a cyclic group is entirely determined by the image of a generator for that group.
$endgroup$
– ÍgjøgnumMeg
Jan 8 at 8:23
$begingroup$
@ÍgjøgnumMeg I still don't see it. can you explain?
$endgroup$
– vesii
Jan 12 at 14:40
$begingroup$
Is $X^prime$ cyclic? What about $Y^prime$? And how could you describe the isomorphism $X^prime cong Y^prime$? What distinguishes $X$ and $Y$?
$endgroup$
– ÍgjøgnumMeg
Jan 7 at 18:27
$begingroup$
Is $X^prime$ cyclic? What about $Y^prime$? And how could you describe the isomorphism $X^prime cong Y^prime$? What distinguishes $X$ and $Y$?
$endgroup$
– ÍgjøgnumMeg
Jan 7 at 18:27
$begingroup$
@ÍgjøgnumMeg I think they are, but how does it help us?
$endgroup$
– vesii
Jan 8 at 0:45
$begingroup$
@ÍgjøgnumMeg I think they are, but how does it help us?
$endgroup$
– vesii
Jan 8 at 0:45
$begingroup$
A cyclic group is generated by a single element, so the image of a homomorphism out of a cyclic group is entirely determined by the image of a generator for that group.
$endgroup$
– ÍgjøgnumMeg
Jan 8 at 8:23
$begingroup$
A cyclic group is generated by a single element, so the image of a homomorphism out of a cyclic group is entirely determined by the image of a generator for that group.
$endgroup$
– ÍgjøgnumMeg
Jan 8 at 8:23
$begingroup$
@ÍgjøgnumMeg I still don't see it. can you explain?
$endgroup$
– vesii
Jan 12 at 14:40
$begingroup$
@ÍgjøgnumMeg I still don't see it. can you explain?
$endgroup$
– vesii
Jan 12 at 14:40
add a comment |
1 Answer
1
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oldest
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$begingroup$
A group homomorphism is entirely determined by the image of the generators for the group, by the homomorphism property. For instance, if $G$ is a cyclic group with generator $g$
then the powers of $g$ determine the elements of $G$. If I now look at a homomorphism $varphi : G to H$ I need only specify the image of $g$ under $varphi$ and this determines the homomorphism because $varphi(g^k) = varphi(g)^k$.
E.g. Let $G = Bbb Z/(5)$ and $H = Bbb Z/(10)$. Then $G$ is a cyclic group generated by $1$. Consider the following map
$$varphi : G to H : 1 mapsto 2.$$
Then $varphi(2) = varphi(1 + 1) = varphi(1) + varphi(1) = 2 + 2 = 4$, etc.
For your question, the implication is that $X^prime$ is considered as a subgroup of $X$ and $Y^prime$ as a subgroup of $Y$ (which you can check of course, free exercise). How should one check that $X^prime cong Y^prime$? Just write down an isomorphism (after checking that $X^prime$ is a cyclic subgroup of $X$).
To check that $X/X^prime cong Y/Y^prime$ you need to know what these quotients are, which should remind you to check that $X^prime triangleleft X$ and $Y^prime triangleleft Y$. Then write down another isomorphism $X/X^prime xrightarrow{sim} Y/Y^prime$.
For the final question, note that orders are preserved by homomorphisms (under the homomorphism property). If $X$ is cyclic then $Y$ should at least have one element with order $lvert X rvert$. Is this the case here?
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
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add a comment |
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$begingroup$
A group homomorphism is entirely determined by the image of the generators for the group, by the homomorphism property. For instance, if $G$ is a cyclic group with generator $g$
then the powers of $g$ determine the elements of $G$. If I now look at a homomorphism $varphi : G to H$ I need only specify the image of $g$ under $varphi$ and this determines the homomorphism because $varphi(g^k) = varphi(g)^k$.
E.g. Let $G = Bbb Z/(5)$ and $H = Bbb Z/(10)$. Then $G$ is a cyclic group generated by $1$. Consider the following map
$$varphi : G to H : 1 mapsto 2.$$
Then $varphi(2) = varphi(1 + 1) = varphi(1) + varphi(1) = 2 + 2 = 4$, etc.
For your question, the implication is that $X^prime$ is considered as a subgroup of $X$ and $Y^prime$ as a subgroup of $Y$ (which you can check of course, free exercise). How should one check that $X^prime cong Y^prime$? Just write down an isomorphism (after checking that $X^prime$ is a cyclic subgroup of $X$).
To check that $X/X^prime cong Y/Y^prime$ you need to know what these quotients are, which should remind you to check that $X^prime triangleleft X$ and $Y^prime triangleleft Y$. Then write down another isomorphism $X/X^prime xrightarrow{sim} Y/Y^prime$.
For the final question, note that orders are preserved by homomorphisms (under the homomorphism property). If $X$ is cyclic then $Y$ should at least have one element with order $lvert X rvert$. Is this the case here?
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
$endgroup$
add a comment |
$begingroup$
A group homomorphism is entirely determined by the image of the generators for the group, by the homomorphism property. For instance, if $G$ is a cyclic group with generator $g$
then the powers of $g$ determine the elements of $G$. If I now look at a homomorphism $varphi : G to H$ I need only specify the image of $g$ under $varphi$ and this determines the homomorphism because $varphi(g^k) = varphi(g)^k$.
E.g. Let $G = Bbb Z/(5)$ and $H = Bbb Z/(10)$. Then $G$ is a cyclic group generated by $1$. Consider the following map
$$varphi : G to H : 1 mapsto 2.$$
Then $varphi(2) = varphi(1 + 1) = varphi(1) + varphi(1) = 2 + 2 = 4$, etc.
For your question, the implication is that $X^prime$ is considered as a subgroup of $X$ and $Y^prime$ as a subgroup of $Y$ (which you can check of course, free exercise). How should one check that $X^prime cong Y^prime$? Just write down an isomorphism (after checking that $X^prime$ is a cyclic subgroup of $X$).
To check that $X/X^prime cong Y/Y^prime$ you need to know what these quotients are, which should remind you to check that $X^prime triangleleft X$ and $Y^prime triangleleft Y$. Then write down another isomorphism $X/X^prime xrightarrow{sim} Y/Y^prime$.
For the final question, note that orders are preserved by homomorphisms (under the homomorphism property). If $X$ is cyclic then $Y$ should at least have one element with order $lvert X rvert$. Is this the case here?
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
$endgroup$
add a comment |
$begingroup$
A group homomorphism is entirely determined by the image of the generators for the group, by the homomorphism property. For instance, if $G$ is a cyclic group with generator $g$
then the powers of $g$ determine the elements of $G$. If I now look at a homomorphism $varphi : G to H$ I need only specify the image of $g$ under $varphi$ and this determines the homomorphism because $varphi(g^k) = varphi(g)^k$.
E.g. Let $G = Bbb Z/(5)$ and $H = Bbb Z/(10)$. Then $G$ is a cyclic group generated by $1$. Consider the following map
$$varphi : G to H : 1 mapsto 2.$$
Then $varphi(2) = varphi(1 + 1) = varphi(1) + varphi(1) = 2 + 2 = 4$, etc.
For your question, the implication is that $X^prime$ is considered as a subgroup of $X$ and $Y^prime$ as a subgroup of $Y$ (which you can check of course, free exercise). How should one check that $X^prime cong Y^prime$? Just write down an isomorphism (after checking that $X^prime$ is a cyclic subgroup of $X$).
To check that $X/X^prime cong Y/Y^prime$ you need to know what these quotients are, which should remind you to check that $X^prime triangleleft X$ and $Y^prime triangleleft Y$. Then write down another isomorphism $X/X^prime xrightarrow{sim} Y/Y^prime$.
For the final question, note that orders are preserved by homomorphisms (under the homomorphism property). If $X$ is cyclic then $Y$ should at least have one element with order $lvert X rvert$. Is this the case here?
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
$endgroup$
A group homomorphism is entirely determined by the image of the generators for the group, by the homomorphism property. For instance, if $G$ is a cyclic group with generator $g$
then the powers of $g$ determine the elements of $G$. If I now look at a homomorphism $varphi : G to H$ I need only specify the image of $g$ under $varphi$ and this determines the homomorphism because $varphi(g^k) = varphi(g)^k$.
E.g. Let $G = Bbb Z/(5)$ and $H = Bbb Z/(10)$. Then $G$ is a cyclic group generated by $1$. Consider the following map
$$varphi : G to H : 1 mapsto 2.$$
Then $varphi(2) = varphi(1 + 1) = varphi(1) + varphi(1) = 2 + 2 = 4$, etc.
For your question, the implication is that $X^prime$ is considered as a subgroup of $X$ and $Y^prime$ as a subgroup of $Y$ (which you can check of course, free exercise). How should one check that $X^prime cong Y^prime$? Just write down an isomorphism (after checking that $X^prime$ is a cyclic subgroup of $X$).
To check that $X/X^prime cong Y/Y^prime$ you need to know what these quotients are, which should remind you to check that $X^prime triangleleft X$ and $Y^prime triangleleft Y$. Then write down another isomorphism $X/X^prime xrightarrow{sim} Y/Y^prime$.
For the final question, note that orders are preserved by homomorphisms (under the homomorphism property). If $X$ is cyclic then $Y$ should at least have one element with order $lvert X rvert$. Is this the case here?
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
answered Jan 13 at 11:40
ÍgjøgnumMegÍgjøgnumMeg
2,84511029
2,84511029
add a comment |
add a comment |
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$begingroup$
Is $X^prime$ cyclic? What about $Y^prime$? And how could you describe the isomorphism $X^prime cong Y^prime$? What distinguishes $X$ and $Y$?
$endgroup$
– ÍgjøgnumMeg
Jan 7 at 18:27
$begingroup$
@ÍgjøgnumMeg I think they are, but how does it help us?
$endgroup$
– vesii
Jan 8 at 0:45
$begingroup$
A cyclic group is generated by a single element, so the image of a homomorphism out of a cyclic group is entirely determined by the image of a generator for that group.
$endgroup$
– ÍgjøgnumMeg
Jan 8 at 8:23
$begingroup$
@ÍgjøgnumMeg I still don't see it. can you explain?
$endgroup$
– vesii
Jan 12 at 14:40