Statement : Every group of order 6 is cyclic - Proof that the statement is false
$begingroup$
Proof
Let G be a group of order 6. By Lagrange's Theorem, G has subgroups of order 1,2,3 and 6.
The subgroups of orders 2 and 3 have prime orders and are cyclic therefor. The subgroup contains an element g of order 2 and the subgroup contains an element h of order 3.
So therefore G is cyclic.
Is this correct or is there also proof of group G not being cyclic
abstract-algebra group-theory cyclic-groups
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
$endgroup$
add a comment |
$begingroup$
Proof
Let G be a group of order 6. By Lagrange's Theorem, G has subgroups of order 1,2,3 and 6.
The subgroups of orders 2 and 3 have prime orders and are cyclic therefor. The subgroup contains an element g of order 2 and the subgroup contains an element h of order 3.
So therefore G is cyclic.
Is this correct or is there also proof of group G not being cyclic
abstract-algebra group-theory cyclic-groups
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
$endgroup$
$begingroup$
Not true, try $D_3$ the dihedral group
$endgroup$
– user25959
Dec 3 '18 at 1:47
$begingroup$
"he subgroup contains an element g of order 2 and the subgroup contains an element h of order 3." but that doesn't mean the order of $hg$ is $6$ or that any element has an order of $6$.
$endgroup$
– fleablood
Dec 3 '18 at 2:03
add a comment |
$begingroup$
Proof
Let G be a group of order 6. By Lagrange's Theorem, G has subgroups of order 1,2,3 and 6.
The subgroups of orders 2 and 3 have prime orders and are cyclic therefor. The subgroup contains an element g of order 2 and the subgroup contains an element h of order 3.
So therefore G is cyclic.
Is this correct or is there also proof of group G not being cyclic
abstract-algebra group-theory cyclic-groups
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
$endgroup$
Proof
Let G be a group of order 6. By Lagrange's Theorem, G has subgroups of order 1,2,3 and 6.
The subgroups of orders 2 and 3 have prime orders and are cyclic therefor. The subgroup contains an element g of order 2 and the subgroup contains an element h of order 3.
So therefore G is cyclic.
Is this correct or is there also proof of group G not being cyclic
abstract-algebra group-theory cyclic-groups
abstract-algebra group-theory cyclic-groups
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
edited Dec 3 '18 at 4:26
Kemono Chen
3,0771743
3,0771743
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
asked Dec 3 '18 at 1:43
winterdeurwinterdeur
62
62
$begingroup$
Not true, try $D_3$ the dihedral group
$endgroup$
– user25959
Dec 3 '18 at 1:47
$begingroup$
"he subgroup contains an element g of order 2 and the subgroup contains an element h of order 3." but that doesn't mean the order of $hg$ is $6$ or that any element has an order of $6$.
$endgroup$
– fleablood
Dec 3 '18 at 2:03
add a comment |
$begingroup$
Not true, try $D_3$ the dihedral group
$endgroup$
– user25959
Dec 3 '18 at 1:47
$begingroup$
"he subgroup contains an element g of order 2 and the subgroup contains an element h of order 3." but that doesn't mean the order of $hg$ is $6$ or that any element has an order of $6$.
$endgroup$
– fleablood
Dec 3 '18 at 2:03
$begingroup$
Not true, try $D_3$ the dihedral group
$endgroup$
– user25959
Dec 3 '18 at 1:47
$begingroup$
Not true, try $D_3$ the dihedral group
$endgroup$
– user25959
Dec 3 '18 at 1:47
$begingroup$
"he subgroup contains an element g of order 2 and the subgroup contains an element h of order 3." but that doesn't mean the order of $hg$ is $6$ or that any element has an order of $6$.
$endgroup$
– fleablood
Dec 3 '18 at 2:03
$begingroup$
"he subgroup contains an element g of order 2 and the subgroup contains an element h of order 3." but that doesn't mean the order of $hg$ is $6$ or that any element has an order of $6$.
$endgroup$
– fleablood
Dec 3 '18 at 2:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To answer your question, it is not correct. Several mistakes here.
Lagrange does not say that such subgroups exist. You are incorrectly applying a false converse. Any subgroup of a group of order $6$ must have order $1, 2, 3$, or $6$. This does not say that such subgroups must exist.
Also, a group $G$ of order $6$ having subgroups that are cyclic does NOT mean that the group $G$ itself is cyclic. In fact, all groups have cyclic subgroups by definition (and usually plenty of them).
If you think the statement is false, then your job is to find a group of order $6$ that is not cyclic. If you think it is true then you must prove that all such groups are cyclic. How many groups of order $6$ do you know?
Edit: never mind, answer given away in the comments.
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
$endgroup$
$begingroup$
Thank you for your comment - I will try to continue as I was clearly wrongly stuck in my own Lagrange theorem
$endgroup$
– winterdeur
Dec 3 '18 at 1:49
$begingroup$
I guess with order 2 I can proof that group G itself is cyclic.
$endgroup$
– winterdeur
Dec 3 '18 at 1:50
$begingroup$
To see a small example to illustrate your mistakes more simply, notice that all proper subgroups of $G=mathbb{Z}_2 times mathbb{Z}_2$ are cyclic, yet $G$ is not. (Your mistakes are also common, so don't worry too much now. You are right to try and fix them now.)
$endgroup$
– Randall
Dec 3 '18 at 1:51
$begingroup$
"I guess with order 2 I can proof that group G itself is cyclic. " Which gets you nowhere. The subgroup of order 2 is cyclic but that doesn't mean the group of which it is a subgroup is cyclic.
$endgroup$
– fleablood
Dec 3 '18 at 1:57
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%2f3023488%2fstatement-every-group-of-order-6-is-cyclic-proof-that-the-statement-is-false%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
$begingroup$
To answer your question, it is not correct. Several mistakes here.
Lagrange does not say that such subgroups exist. You are incorrectly applying a false converse. Any subgroup of a group of order $6$ must have order $1, 2, 3$, or $6$. This does not say that such subgroups must exist.
Also, a group $G$ of order $6$ having subgroups that are cyclic does NOT mean that the group $G$ itself is cyclic. In fact, all groups have cyclic subgroups by definition (and usually plenty of them).
If you think the statement is false, then your job is to find a group of order $6$ that is not cyclic. If you think it is true then you must prove that all such groups are cyclic. How many groups of order $6$ do you know?
Edit: never mind, answer given away in the comments.
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
$endgroup$
$begingroup$
Thank you for your comment - I will try to continue as I was clearly wrongly stuck in my own Lagrange theorem
$endgroup$
– winterdeur
Dec 3 '18 at 1:49
$begingroup$
I guess with order 2 I can proof that group G itself is cyclic.
$endgroup$
– winterdeur
Dec 3 '18 at 1:50
$begingroup$
To see a small example to illustrate your mistakes more simply, notice that all proper subgroups of $G=mathbb{Z}_2 times mathbb{Z}_2$ are cyclic, yet $G$ is not. (Your mistakes are also common, so don't worry too much now. You are right to try and fix them now.)
$endgroup$
– Randall
Dec 3 '18 at 1:51
$begingroup$
"I guess with order 2 I can proof that group G itself is cyclic. " Which gets you nowhere. The subgroup of order 2 is cyclic but that doesn't mean the group of which it is a subgroup is cyclic.
$endgroup$
– fleablood
Dec 3 '18 at 1:57
add a comment |
$begingroup$
To answer your question, it is not correct. Several mistakes here.
Lagrange does not say that such subgroups exist. You are incorrectly applying a false converse. Any subgroup of a group of order $6$ must have order $1, 2, 3$, or $6$. This does not say that such subgroups must exist.
Also, a group $G$ of order $6$ having subgroups that are cyclic does NOT mean that the group $G$ itself is cyclic. In fact, all groups have cyclic subgroups by definition (and usually plenty of them).
If you think the statement is false, then your job is to find a group of order $6$ that is not cyclic. If you think it is true then you must prove that all such groups are cyclic. How many groups of order $6$ do you know?
Edit: never mind, answer given away in the comments.
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
$endgroup$
$begingroup$
Thank you for your comment - I will try to continue as I was clearly wrongly stuck in my own Lagrange theorem
$endgroup$
– winterdeur
Dec 3 '18 at 1:49
$begingroup$
I guess with order 2 I can proof that group G itself is cyclic.
$endgroup$
– winterdeur
Dec 3 '18 at 1:50
$begingroup$
To see a small example to illustrate your mistakes more simply, notice that all proper subgroups of $G=mathbb{Z}_2 times mathbb{Z}_2$ are cyclic, yet $G$ is not. (Your mistakes are also common, so don't worry too much now. You are right to try and fix them now.)
$endgroup$
– Randall
Dec 3 '18 at 1:51
$begingroup$
"I guess with order 2 I can proof that group G itself is cyclic. " Which gets you nowhere. The subgroup of order 2 is cyclic but that doesn't mean the group of which it is a subgroup is cyclic.
$endgroup$
– fleablood
Dec 3 '18 at 1:57
add a comment |
$begingroup$
To answer your question, it is not correct. Several mistakes here.
Lagrange does not say that such subgroups exist. You are incorrectly applying a false converse. Any subgroup of a group of order $6$ must have order $1, 2, 3$, or $6$. This does not say that such subgroups must exist.
Also, a group $G$ of order $6$ having subgroups that are cyclic does NOT mean that the group $G$ itself is cyclic. In fact, all groups have cyclic subgroups by definition (and usually plenty of them).
If you think the statement is false, then your job is to find a group of order $6$ that is not cyclic. If you think it is true then you must prove that all such groups are cyclic. How many groups of order $6$ do you know?
Edit: never mind, answer given away in the comments.
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
$endgroup$
To answer your question, it is not correct. Several mistakes here.
Lagrange does not say that such subgroups exist. You are incorrectly applying a false converse. Any subgroup of a group of order $6$ must have order $1, 2, 3$, or $6$. This does not say that such subgroups must exist.
Also, a group $G$ of order $6$ having subgroups that are cyclic does NOT mean that the group $G$ itself is cyclic. In fact, all groups have cyclic subgroups by definition (and usually plenty of them).
If you think the statement is false, then your job is to find a group of order $6$ that is not cyclic. If you think it is true then you must prove that all such groups are cyclic. How many groups of order $6$ do you know?
Edit: never mind, answer given away in the comments.
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
answered Dec 3 '18 at 1:47
RandallRandall
9,94111230
9,94111230
$begingroup$
Thank you for your comment - I will try to continue as I was clearly wrongly stuck in my own Lagrange theorem
$endgroup$
– winterdeur
Dec 3 '18 at 1:49
$begingroup$
I guess with order 2 I can proof that group G itself is cyclic.
$endgroup$
– winterdeur
Dec 3 '18 at 1:50
$begingroup$
To see a small example to illustrate your mistakes more simply, notice that all proper subgroups of $G=mathbb{Z}_2 times mathbb{Z}_2$ are cyclic, yet $G$ is not. (Your mistakes are also common, so don't worry too much now. You are right to try and fix them now.)
$endgroup$
– Randall
Dec 3 '18 at 1:51
$begingroup$
"I guess with order 2 I can proof that group G itself is cyclic. " Which gets you nowhere. The subgroup of order 2 is cyclic but that doesn't mean the group of which it is a subgroup is cyclic.
$endgroup$
– fleablood
Dec 3 '18 at 1:57
add a comment |
$begingroup$
Thank you for your comment - I will try to continue as I was clearly wrongly stuck in my own Lagrange theorem
$endgroup$
– winterdeur
Dec 3 '18 at 1:49
$begingroup$
I guess with order 2 I can proof that group G itself is cyclic.
$endgroup$
– winterdeur
Dec 3 '18 at 1:50
$begingroup$
To see a small example to illustrate your mistakes more simply, notice that all proper subgroups of $G=mathbb{Z}_2 times mathbb{Z}_2$ are cyclic, yet $G$ is not. (Your mistakes are also common, so don't worry too much now. You are right to try and fix them now.)
$endgroup$
– Randall
Dec 3 '18 at 1:51
$begingroup$
"I guess with order 2 I can proof that group G itself is cyclic. " Which gets you nowhere. The subgroup of order 2 is cyclic but that doesn't mean the group of which it is a subgroup is cyclic.
$endgroup$
– fleablood
Dec 3 '18 at 1:57
$begingroup$
Thank you for your comment - I will try to continue as I was clearly wrongly stuck in my own Lagrange theorem
$endgroup$
– winterdeur
Dec 3 '18 at 1:49
$begingroup$
Thank you for your comment - I will try to continue as I was clearly wrongly stuck in my own Lagrange theorem
$endgroup$
– winterdeur
Dec 3 '18 at 1:49
$begingroup$
I guess with order 2 I can proof that group G itself is cyclic.
$endgroup$
– winterdeur
Dec 3 '18 at 1:50
$begingroup$
I guess with order 2 I can proof that group G itself is cyclic.
$endgroup$
– winterdeur
Dec 3 '18 at 1:50
$begingroup$
To see a small example to illustrate your mistakes more simply, notice that all proper subgroups of $G=mathbb{Z}_2 times mathbb{Z}_2$ are cyclic, yet $G$ is not. (Your mistakes are also common, so don't worry too much now. You are right to try and fix them now.)
$endgroup$
– Randall
Dec 3 '18 at 1:51
$begingroup$
To see a small example to illustrate your mistakes more simply, notice that all proper subgroups of $G=mathbb{Z}_2 times mathbb{Z}_2$ are cyclic, yet $G$ is not. (Your mistakes are also common, so don't worry too much now. You are right to try and fix them now.)
$endgroup$
– Randall
Dec 3 '18 at 1:51
$begingroup$
"I guess with order 2 I can proof that group G itself is cyclic. " Which gets you nowhere. The subgroup of order 2 is cyclic but that doesn't mean the group of which it is a subgroup is cyclic.
$endgroup$
– fleablood
Dec 3 '18 at 1:57
$begingroup$
"I guess with order 2 I can proof that group G itself is cyclic. " Which gets you nowhere. The subgroup of order 2 is cyclic but that doesn't mean the group of which it is a subgroup is cyclic.
$endgroup$
– fleablood
Dec 3 '18 at 1:57
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%2f3023488%2fstatement-every-group-of-order-6-is-cyclic-proof-that-the-statement-is-false%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
$begingroup$
Not true, try $D_3$ the dihedral group
$endgroup$
– user25959
Dec 3 '18 at 1:47
$begingroup$
"he subgroup contains an element g of order 2 and the subgroup contains an element h of order 3." but that doesn't mean the order of $hg$ is $6$ or that any element has an order of $6$.
$endgroup$
– fleablood
Dec 3 '18 at 2:03