$G$ be a group where $G= mathbb{Z}_6 oplus mathbb{Z}_8$ and the normal subgroup $H=langle(2,4)rangle$ use...
$begingroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
$endgroup$
add a comment |
$begingroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
$endgroup$
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
add a comment |
$begingroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
$endgroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
1
asked Dec 5 '18 at 3:31
user595420
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
add a comment |
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
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%2f3026569%2fg-be-a-group-where-g-mathbbz-6-oplus-mathbbz-8-and-the-normal-subgro%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
$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
add a comment |
$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
add a comment |
$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
edited Dec 5 '18 at 5:57
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
13.9k3827
add a comment |
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%2f3026569%2fg-be-a-group-where-g-mathbbz-6-oplus-mathbbz-8-and-the-normal-subgro%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
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26