Statement : Every group of order 6 is cyclic - Proof that the statement is false












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 4:26









Kemono Chen

2,9181739




2,9181739










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

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2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 '18 at 1:47









RandallRandall

9,42111129




9,42111129












  • $begingroup$
    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










  • $begingroup$
    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


















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