Mixing
Understanding the group product of $mathbb{Z_n}$ and $mathbb{Z_m}$, $G=mathbb{Z_m} mathbb{Z_n}$
$begingroup$
$G=mathbb{Z_m} mathbb{Z_n} = {[a][b] : [a] in mathbb{Z_m}, [b] in mathbb{Z_n}}$
Specifically when $gcd(m,n)=1$. Can somebody show me what $G$ will equal as a set, and if you could go to far as to open up the equivalence classes of $[a],[b]$ and show me what exactly is going on, I'd appreciate it. I know that when $gcd(m,n)=1$ this product will be isomorphic to $mathbb{Z_{mn}}$ and am just looking for a bit of details to clarify. Thanks!
group-theory number-theory finite-groups cyclic-groups
edited Feb 2 at 17:44
Shaun
10.5k113687
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
$endgroup$
add a comment |
$begingroup$
$G=mathbb{Z_m} mathbb{Z_n} = {[a][b] : [a] in mathbb{Z_m}, [b] in mathbb{Z_n}}$
Specifically when $gcd(m,n)=1$. Can somebody show me what $G$ will equal as a set, and if you could go to far as to open up the equivalence classes of $[a],[b]$ and show me what exactly is going on, I'd appreciate it. I know that when $gcd(m,n)=1$ this product will be isomorphic to $mathbb{Z_{mn}}$ and am just looking for a bit of details to clarify. Thanks!
group-theory number-theory finite-groups cyclic-groups
edited Feb 2 at 17:44
Shaun
10.5k113687
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
$endgroup$
1
$begingroup$
Do you mean the direct product $G=mathbb{Z}_m times mathbb{Z}_n$ ?
$endgroup$
– lhf
Feb 1 at 18:08
$begingroup$
No, I don't actually. Like, if you have two subgroups of $G$, say $H,K$, then you can describe $HK$ as a set, and this set is a subgroup of $G$ if either $K$ or $H$ is normal in $G$. I'm looking for someone to explain to me why $mathbb{Z_n} mathbb{Z_m} = mathbb{Z_{mn}}$ when $gcd(m,n)=1$, and specifically I would like to "open up the equivalence classes" and see why this makes sense in a number theoretic sort of way
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:17
1
$begingroup$
So $mathbb Z_n$, $mathbb Z_m$ are subgroups of which group?
$endgroup$
– eduard
Feb 1 at 18:24
$begingroup$
Ah, right, sorry. They are isomorphic to subgroups of $mathbb{Z_n} times mathbb{Z_m}$
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:33
$begingroup$
Can you explain your attempt? Do you feel comfortable with Isomorphism theorems?
$endgroup$
– eduard
Feb 1 at 18:58
add a comment |
$begingroup$
$G=mathbb{Z_m} mathbb{Z_n} = {[a][b] : [a] in mathbb{Z_m}, [b] in mathbb{Z_n}}$
Specifically when $gcd(m,n)=1$. Can somebody show me what $G$ will equal as a set, and if you could go to far as to open up the equivalence classes of $[a],[b]$ and show me what exactly is going on, I'd appreciate it. I know that when $gcd(m,n)=1$ this product will be isomorphic to $mathbb{Z_{mn}}$ and am just looking for a bit of details to clarify. Thanks!
group-theory number-theory finite-groups cyclic-groups
edited Feb 2 at 17:44
Shaun
10.5k113687
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
$endgroup$
$G=mathbb{Z_m} mathbb{Z_n} = {[a][b] : [a] in mathbb{Z_m}, [b] in mathbb{Z_n}}$
Specifically when $gcd(m,n)=1$. Can somebody show me what $G$ will equal as a set, and if you could go to far as to open up the equivalence classes of $[a],[b]$ and show me what exactly is going on, I'd appreciate it. I know that when $gcd(m,n)=1$ this product will be isomorphic to $mathbb{Z_{mn}}$ and am just looking for a bit of details to clarify. Thanks!
group-theory number-theory finite-groups cyclic-groups
group-theory number-theory finite-groups cyclic-groups
edited Feb 2 at 17:44
Shaun
10.5k113687
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
edited Feb 2 at 17:44
Shaun
10.5k113687
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
edited Feb 2 at 17:44
Shaun
10.5k113687
edited Feb 2 at 17:44
Shaun
10.5k113687
edited Feb 2 at 17:44
Shaun
10.5k113687
10.5k113687
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
asked Feb 1 at 18:04
Mathematical MushroomMathematical Mushroom
1178
1178
1
$begingroup$
Do you mean the direct product $G=mathbb{Z}_m times mathbb{Z}_n$ ?
$endgroup$
– lhf
Feb 1 at 18:08
$begingroup$
No, I don't actually. Like, if you have two subgroups of $G$, say $H,K$, then you can describe $HK$ as a set, and this set is a subgroup of $G$ if either $K$ or $H$ is normal in $G$. I'm looking for someone to explain to me why $mathbb{Z_n} mathbb{Z_m} = mathbb{Z_{mn}}$ when $gcd(m,n)=1$, and specifically I would like to "open up the equivalence classes" and see why this makes sense in a number theoretic sort of way
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:17
1
$begingroup$
So $mathbb Z_n$, $mathbb Z_m$ are subgroups of which group?
$endgroup$
– eduard
Feb 1 at 18:24
$begingroup$
Ah, right, sorry. They are isomorphic to subgroups of $mathbb{Z_n} times mathbb{Z_m}$
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:33
$begingroup$
Can you explain your attempt? Do you feel comfortable with Isomorphism theorems?
$endgroup$
– eduard
Feb 1 at 18:58
add a comment |
1
$begingroup$
Do you mean the direct product $G=mathbb{Z}_m times mathbb{Z}_n$ ?
$endgroup$
– lhf
Feb 1 at 18:08
$begingroup$
No, I don't actually. Like, if you have two subgroups of $G$, say $H,K$, then you can describe $HK$ as a set, and this set is a subgroup of $G$ if either $K$ or $H$ is normal in $G$. I'm looking for someone to explain to me why $mathbb{Z_n} mathbb{Z_m} = mathbb{Z_{mn}}$ when $gcd(m,n)=1$, and specifically I would like to "open up the equivalence classes" and see why this makes sense in a number theoretic sort of way
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:17
1
$begingroup$
So $mathbb Z_n$, $mathbb Z_m$ are subgroups of which group?
$endgroup$
– eduard
Feb 1 at 18:24
$begingroup$
Ah, right, sorry. They are isomorphic to subgroups of $mathbb{Z_n} times mathbb{Z_m}$
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:33
$begingroup$
Can you explain your attempt? Do you feel comfortable with Isomorphism theorems?
$endgroup$
– eduard
Feb 1 at 18:58
1
1
$begingroup$
Do you mean the direct product $G=mathbb{Z}_m times mathbb{Z}_n$ ?
$endgroup$
– lhf
Feb 1 at 18:08
$begingroup$
Do you mean the direct product $G=mathbb{Z}_m times mathbb{Z}_n$ ?
$endgroup$
– lhf
Feb 1 at 18:08
$begingroup$
No, I don't actually. Like, if you have two subgroups of $G$, say $H,K$, then you can describe $HK$ as a set, and this set is a subgroup of $G$ if either $K$ or $H$ is normal in $G$. I'm looking for someone to explain to me why $mathbb{Z_n} mathbb{Z_m} = mathbb{Z_{mn}}$ when $gcd(m,n)=1$, and specifically I would like to "open up the equivalence classes" and see why this makes sense in a number theoretic sort of way
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:17
$begingroup$
No, I don't actually. Like, if you have two subgroups of $G$, say $H,K$, then you can describe $HK$ as a set, and this set is a subgroup of $G$ if either $K$ or $H$ is normal in $G$. I'm looking for someone to explain to me why $mathbb{Z_n} mathbb{Z_m} = mathbb{Z_{mn}}$ when $gcd(m,n)=1$, and specifically I would like to "open up the equivalence classes" and see why this makes sense in a number theoretic sort of way
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:17
1
1
$begingroup$
So $mathbb Z_n$, $mathbb Z_m$ are subgroups of which group?
$endgroup$
– eduard
Feb 1 at 18:24
$begingroup$
So $mathbb Z_n$, $mathbb Z_m$ are subgroups of which group?
$endgroup$
– eduard
Feb 1 at 18:24
$begingroup$
Ah, right, sorry. They are isomorphic to subgroups of $mathbb{Z_n} times mathbb{Z_m}$
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:33
$begingroup$
Ah, right, sorry. They are isomorphic to subgroups of $mathbb{Z_n} times mathbb{Z_m}$
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:33
$begingroup$
Can you explain your attempt? Do you feel comfortable with Isomorphism theorems?
$endgroup$
– eduard
Feb 1 at 18:58
$begingroup$
Can you explain your attempt? Do you feel comfortable with Isomorphism theorems?
$endgroup$
– eduard
Feb 1 at 18:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Algebraically it would most likely be easier to start with a group $H = mathbb{Z}_n times mathbb{Z}_m$ with $gcd(m,n) = 1$. Then it's very straightforward to see that $mathbb{Z}_n mathbb{Z}_m cong H$ (and many proofs of the Chinese Remainder Theorem include this line of reasoning; recall the Chinese Remainder Theorem says essentially that $mathbb{Z}_n times mathbb{Z}_m cong mathbb{Z}_{nm}$ exactly when $gcd(m,n) = 1$).
But you seem to want a more hand-on understanding. So let's produce an example, say with $6 = 2 cdot 3$. And we'll begin by starting with the group $H = mathbb{Z}_6$, which I'll recognize as $(mathbb{Z}/7mathbb{Z})^times$, the multiplicative units mod $7$.
(As a small note: writing $[n]$ for equivalence classes all the time is annoying. But essentially integer that appears below that is a member of a group is an equivalence class in that group).
Let $A = langle 2 rangle = { 2, 4, 1 } cong mathbb{Z}_3$ be the subgroup generated by $2$ and let $B = langle 6 rangle = {6, 1 } cong mathbb{Z}_2$ be the subgroup generated by $6$.
Then your question asks what $AB$ looks like as a subset of $H$. Of course, we know that $AB = H$, so really I suppose we're asking if we can gain an understanding of how $AB = H$.
So let's do it. The product $AB$ consists of the following $6$ elements (coming from $3$ choices from $A$ and $2$ choices from $B$):
$$begin{array}{lllll}
2 cdot 6 = 5 && 4 cdot 6 = 3 && 1 cdot 6 = 6 \
2 cdot 1 = 2 && 4 cdot 1 = 4 && 1 cdot 1 = 1
end{array}$$
Notice that each element $1, 2, 3, 4, 5, 6$ is represented, and thus (as sets) it's clear that $AB = H$. From this it's immediately clear that this is true as groups as well, knowing that a subgroup of $H$ with the same (finite) size as $H$ must be $H$.
Further, note that each element is represented in a unique way! This is also part of the statement of the Chinese Remainder Theorem.
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
$begingroup$
Thanks. I wanted to see that $mathbb{Z_n} mathbb{Z_m}=mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $mathbb{Z_n} times mathbb{Z_m}=mathbb{Z_{mn}}$
$endgroup$
– Mathematical Mushroom
Feb 4 at 1:29
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096548%2funderstanding-the-group-product-of-mathbbz-n-and-mathbbz-m-g-mathbb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Algebraically it would most likely be easier to start with a group $H = mathbb{Z}_n times mathbb{Z}_m$ with $gcd(m,n) = 1$. Then it's very straightforward to see that $mathbb{Z}_n mathbb{Z}_m cong H$ (and many proofs of the Chinese Remainder Theorem include this line of reasoning; recall the Chinese Remainder Theorem says essentially that $mathbb{Z}_n times mathbb{Z}_m cong mathbb{Z}_{nm}$ exactly when $gcd(m,n) = 1$).
But you seem to want a more hand-on understanding. So let's produce an example, say with $6 = 2 cdot 3$. And we'll begin by starting with the group $H = mathbb{Z}_6$, which I'll recognize as $(mathbb{Z}/7mathbb{Z})^times$, the multiplicative units mod $7$.
(As a small note: writing $[n]$ for equivalence classes all the time is annoying. But essentially integer that appears below that is a member of a group is an equivalence class in that group).
Let $A = langle 2 rangle = { 2, 4, 1 } cong mathbb{Z}_3$ be the subgroup generated by $2$ and let $B = langle 6 rangle = {6, 1 } cong mathbb{Z}_2$ be the subgroup generated by $6$.
Then your question asks what $AB$ looks like as a subset of $H$. Of course, we know that $AB = H$, so really I suppose we're asking if we can gain an understanding of how $AB = H$.
So let's do it. The product $AB$ consists of the following $6$ elements (coming from $3$ choices from $A$ and $2$ choices from $B$):
$$begin{array}{lllll}
2 cdot 6 = 5 && 4 cdot 6 = 3 && 1 cdot 6 = 6 \
2 cdot 1 = 2 && 4 cdot 1 = 4 && 1 cdot 1 = 1
end{array}$$
Notice that each element $1, 2, 3, 4, 5, 6$ is represented, and thus (as sets) it's clear that $AB = H$. From this it's immediately clear that this is true as groups as well, knowing that a subgroup of $H$ with the same (finite) size as $H$ must be $H$.
Further, note that each element is represented in a unique way! This is also part of the statement of the Chinese Remainder Theorem.
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
$begingroup$
Thanks. I wanted to see that $mathbb{Z_n} mathbb{Z_m}=mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $mathbb{Z_n} times mathbb{Z_m}=mathbb{Z_{mn}}$
$endgroup$
– Mathematical Mushroom
Feb 4 at 1:29
add a comment |
$begingroup$
Algebraically it would most likely be easier to start with a group $H = mathbb{Z}_n times mathbb{Z}_m$ with $gcd(m,n) = 1$. Then it's very straightforward to see that $mathbb{Z}_n mathbb{Z}_m cong H$ (and many proofs of the Chinese Remainder Theorem include this line of reasoning; recall the Chinese Remainder Theorem says essentially that $mathbb{Z}_n times mathbb{Z}_m cong mathbb{Z}_{nm}$ exactly when $gcd(m,n) = 1$).
But you seem to want a more hand-on understanding. So let's produce an example, say with $6 = 2 cdot 3$. And we'll begin by starting with the group $H = mathbb{Z}_6$, which I'll recognize as $(mathbb{Z}/7mathbb{Z})^times$, the multiplicative units mod $7$.
(As a small note: writing $[n]$ for equivalence classes all the time is annoying. But essentially integer that appears below that is a member of a group is an equivalence class in that group).
Let $A = langle 2 rangle = { 2, 4, 1 } cong mathbb{Z}_3$ be the subgroup generated by $2$ and let $B = langle 6 rangle = {6, 1 } cong mathbb{Z}_2$ be the subgroup generated by $6$.
Then your question asks what $AB$ looks like as a subset of $H$. Of course, we know that $AB = H$, so really I suppose we're asking if we can gain an understanding of how $AB = H$.
So let's do it. The product $AB$ consists of the following $6$ elements (coming from $3$ choices from $A$ and $2$ choices from $B$):
$$begin{array}{lllll}
2 cdot 6 = 5 && 4 cdot 6 = 3 && 1 cdot 6 = 6 \
2 cdot 1 = 2 && 4 cdot 1 = 4 && 1 cdot 1 = 1
end{array}$$
Notice that each element $1, 2, 3, 4, 5, 6$ is represented, and thus (as sets) it's clear that $AB = H$. From this it's immediately clear that this is true as groups as well, knowing that a subgroup of $H$ with the same (finite) size as $H$ must be $H$.
Further, note that each element is represented in a unique way! This is also part of the statement of the Chinese Remainder Theorem.
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
$begingroup$
Thanks. I wanted to see that $mathbb{Z_n} mathbb{Z_m}=mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $mathbb{Z_n} times mathbb{Z_m}=mathbb{Z_{mn}}$
$endgroup$
– Mathematical Mushroom
Feb 4 at 1:29
add a comment |
$begingroup$
Algebraically it would most likely be easier to start with a group $H = mathbb{Z}_n times mathbb{Z}_m$ with $gcd(m,n) = 1$. Then it's very straightforward to see that $mathbb{Z}_n mathbb{Z}_m cong H$ (and many proofs of the Chinese Remainder Theorem include this line of reasoning; recall the Chinese Remainder Theorem says essentially that $mathbb{Z}_n times mathbb{Z}_m cong mathbb{Z}_{nm}$ exactly when $gcd(m,n) = 1$).
But you seem to want a more hand-on understanding. So let's produce an example, say with $6 = 2 cdot 3$. And we'll begin by starting with the group $H = mathbb{Z}_6$, which I'll recognize as $(mathbb{Z}/7mathbb{Z})^times$, the multiplicative units mod $7$.
(As a small note: writing $[n]$ for equivalence classes all the time is annoying. But essentially integer that appears below that is a member of a group is an equivalence class in that group).
Let $A = langle 2 rangle = { 2, 4, 1 } cong mathbb{Z}_3$ be the subgroup generated by $2$ and let $B = langle 6 rangle = {6, 1 } cong mathbb{Z}_2$ be the subgroup generated by $6$.
Then your question asks what $AB$ looks like as a subset of $H$. Of course, we know that $AB = H$, so really I suppose we're asking if we can gain an understanding of how $AB = H$.
So let's do it. The product $AB$ consists of the following $6$ elements (coming from $3$ choices from $A$ and $2$ choices from $B$):
$$begin{array}{lllll}
2 cdot 6 = 5 && 4 cdot 6 = 3 && 1 cdot 6 = 6 \
2 cdot 1 = 2 && 4 cdot 1 = 4 && 1 cdot 1 = 1
end{array}$$
Notice that each element $1, 2, 3, 4, 5, 6$ is represented, and thus (as sets) it's clear that $AB = H$. From this it's immediately clear that this is true as groups as well, knowing that a subgroup of $H$ with the same (finite) size as $H$ must be $H$.
Further, note that each element is represented in a unique way! This is also part of the statement of the Chinese Remainder Theorem.
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
$endgroup$
Algebraically it would most likely be easier to start with a group $H = mathbb{Z}_n times mathbb{Z}_m$ with $gcd(m,n) = 1$. Then it's very straightforward to see that $mathbb{Z}_n mathbb{Z}_m cong H$ (and many proofs of the Chinese Remainder Theorem include this line of reasoning; recall the Chinese Remainder Theorem says essentially that $mathbb{Z}_n times mathbb{Z}_m cong mathbb{Z}_{nm}$ exactly when $gcd(m,n) = 1$).
But you seem to want a more hand-on understanding. So let's produce an example, say with $6 = 2 cdot 3$. And we'll begin by starting with the group $H = mathbb{Z}_6$, which I'll recognize as $(mathbb{Z}/7mathbb{Z})^times$, the multiplicative units mod $7$.
(As a small note: writing $[n]$ for equivalence classes all the time is annoying. But essentially integer that appears below that is a member of a group is an equivalence class in that group).
Let $A = langle 2 rangle = { 2, 4, 1 } cong mathbb{Z}_3$ be the subgroup generated by $2$ and let $B = langle 6 rangle = {6, 1 } cong mathbb{Z}_2$ be the subgroup generated by $6$.
Then your question asks what $AB$ looks like as a subset of $H$. Of course, we know that $AB = H$, so really I suppose we're asking if we can gain an understanding of how $AB = H$.
So let's do it. The product $AB$ consists of the following $6$ elements (coming from $3$ choices from $A$ and $2$ choices from $B$):
$$begin{array}{lllll}
2 cdot 6 = 5 && 4 cdot 6 = 3 && 1 cdot 6 = 6 \
2 cdot 1 = 2 && 4 cdot 1 = 4 && 1 cdot 1 = 1
end{array}$$
Notice that each element $1, 2, 3, 4, 5, 6$ is represented, and thus (as sets) it's clear that $AB = H$. From this it's immediately clear that this is true as groups as well, knowing that a subgroup of $H$ with the same (finite) size as $H$ must be $H$.
Further, note that each element is represented in a unique way! This is also part of the statement of the Chinese Remainder Theorem.
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
answered Feb 3 at 15:25
davidlowryduda♦davidlowryduda
75.2k7121256
75.2k7121256
$begingroup$
Thanks. I wanted to see that $mathbb{Z_n} mathbb{Z_m}=mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $mathbb{Z_n} times mathbb{Z_m}=mathbb{Z_{mn}}$
$endgroup$
– Mathematical Mushroom
Feb 4 at 1:29
add a comment |
$begingroup$
Thanks. I wanted to see that $mathbb{Z_n} mathbb{Z_m}=mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $mathbb{Z_n} times mathbb{Z_m}=mathbb{Z_{mn}}$
$endgroup$
– Mathematical Mushroom
Feb 4 at 1:29
$begingroup$
Thanks. I wanted to see that $mathbb{Z_n} mathbb{Z_m}=mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $mathbb{Z_n} times mathbb{Z_m}=mathbb{Z_{mn}}$
$endgroup$
– Mathematical Mushroom
Feb 4 at 1:29
$begingroup$
Thanks. I wanted to see that $mathbb{Z_n} mathbb{Z_m}=mathbb{Z_{mn}}$ when $gcd(m,n)=1$ in order to conclude that $mathbb{Z_n} times mathbb{Z_m}=mathbb{Z_{mn}}$
$endgroup$
– Mathematical Mushroom
Feb 4 at 1:29
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096548%2funderstanding-the-group-product-of-mathbbz-n-and-mathbbz-m-g-mathbb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Do you mean the direct product $G=mathbb{Z}_m times mathbb{Z}_n$ ?
$endgroup$
– lhf
Feb 1 at 18:08
$begingroup$
No, I don't actually. Like, if you have two subgroups of $G$, say $H,K$, then you can describe $HK$ as a set, and this set is a subgroup of $G$ if either $K$ or $H$ is normal in $G$. I'm looking for someone to explain to me why $mathbb{Z_n} mathbb{Z_m} = mathbb{Z_{mn}}$ when $gcd(m,n)=1$, and specifically I would like to "open up the equivalence classes" and see why this makes sense in a number theoretic sort of way
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:17
1
$begingroup$
So $mathbb Z_n$, $mathbb Z_m$ are subgroups of which group?
$endgroup$
– eduard
Feb 1 at 18:24
$begingroup$
Ah, right, sorry. They are isomorphic to subgroups of $mathbb{Z_n} times mathbb{Z_m}$
$endgroup$
– Mathematical Mushroom
Feb 1 at 18:33
$begingroup$
Can you explain your attempt? Do you feel comfortable with Isomorphism theorems?
$endgroup$
– eduard
Feb 1 at 18:58