Group under Additive modulo 6

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Is the set $0,1,2,3,4,5,6$ a group under additive modulo $6$?



My Try:



The inverse of this group would be 0.
The Cayley-table entry for 6 would contain 0 at two locations
$6+_60=0$ and $6+_66=0$, but in a group the Cayley table entries are unique!!.
So this set is not a group.



Please let me know if I am correct?




group-theory




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  • 3




    If you work modulo $6$ then $0=6$, so there is a redundancy in the way you wrote the set. Also, you mix up inverse and identity element.
    – Arnaud Mortier
    Jul 15 at 11:51










  • Didn't get you Arnaud :(
    – user3767495
    Jul 15 at 11:53










  • @user3767495 What Arnaud said was that in your group, $0$ is the same element as $6$, hence the redundancy. Second, you could have said that the "identity" of the group is $0$.
    – Frenzy Li
    Jul 15 at 12:06






  • 1




    You are correct in saying that the answer to your question is no.
    – Derek Holt
    Jul 15 at 12:42














up vote
1
down vote

favorite












Is the set $0,1,2,3,4,5,6$ a group under additive modulo $6$?



My Try:



The inverse of this group would be 0.
The Cayley-table entry for 6 would contain 0 at two locations
$6+_60=0$ and $6+_66=0$, but in a group the Cayley table entries are unique!!.
So this set is not a group.



Please let me know if I am correct?




group-theory




share|cite|improve this question

















  • 3




    If you work modulo $6$ then $0=6$, so there is a redundancy in the way you wrote the set. Also, you mix up inverse and identity element.
    – Arnaud Mortier
    Jul 15 at 11:51










  • Didn't get you Arnaud :(
    – user3767495
    Jul 15 at 11:53










  • @user3767495 What Arnaud said was that in your group, $0$ is the same element as $6$, hence the redundancy. Second, you could have said that the "identity" of the group is $0$.
    – Frenzy Li
    Jul 15 at 12:06






  • 1




    You are correct in saying that the answer to your question is no.
    – Derek Holt
    Jul 15 at 12:42












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is the set $0,1,2,3,4,5,6$ a group under additive modulo $6$?



My Try:



The inverse of this group would be 0.
The Cayley-table entry for 6 would contain 0 at two locations
$6+_60=0$ and $6+_66=0$, but in a group the Cayley table entries are unique!!.
So this set is not a group.



Please let me know if I am correct?




group-theory




share|cite|improve this question













Is the set $0,1,2,3,4,5,6$ a group under additive modulo $6$?



My Try:



The inverse of this group would be 0.
The Cayley-table entry for 6 would contain 0 at two locations
$6+_60=0$ and $6+_66=0$, but in a group the Cayley table entries are unique!!.
So this set is not a group.



Please let me know if I am correct?





group-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 15 at 11:52









Key Flex

4,406525




4,406525









asked Jul 15 at 11:49









user3767495

857




857







  • 3




    If you work modulo $6$ then $0=6$, so there is a redundancy in the way you wrote the set. Also, you mix up inverse and identity element.
    – Arnaud Mortier
    Jul 15 at 11:51










  • Didn't get you Arnaud :(
    – user3767495
    Jul 15 at 11:53










  • @user3767495 What Arnaud said was that in your group, $0$ is the same element as $6$, hence the redundancy. Second, you could have said that the "identity" of the group is $0$.
    – Frenzy Li
    Jul 15 at 12:06






  • 1




    You are correct in saying that the answer to your question is no.
    – Derek Holt
    Jul 15 at 12:42












  • 3




    If you work modulo $6$ then $0=6$, so there is a redundancy in the way you wrote the set. Also, you mix up inverse and identity element.
    – Arnaud Mortier
    Jul 15 at 11:51










  • Didn't get you Arnaud :(
    – user3767495
    Jul 15 at 11:53










  • @user3767495 What Arnaud said was that in your group, $0$ is the same element as $6$, hence the redundancy. Second, you could have said that the "identity" of the group is $0$.
    – Frenzy Li
    Jul 15 at 12:06






  • 1




    You are correct in saying that the answer to your question is no.
    – Derek Holt
    Jul 15 at 12:42







3




3




If you work modulo $6$ then $0=6$, so there is a redundancy in the way you wrote the set. Also, you mix up inverse and identity element.
– Arnaud Mortier
Jul 15 at 11:51




If you work modulo $6$ then $0=6$, so there is a redundancy in the way you wrote the set. Also, you mix up inverse and identity element.
– Arnaud Mortier
Jul 15 at 11:51












Didn't get you Arnaud :(
– user3767495
Jul 15 at 11:53




Didn't get you Arnaud :(
– user3767495
Jul 15 at 11:53












@user3767495 What Arnaud said was that in your group, $0$ is the same element as $6$, hence the redundancy. Second, you could have said that the "identity" of the group is $0$.
– Frenzy Li
Jul 15 at 12:06




@user3767495 What Arnaud said was that in your group, $0$ is the same element as $6$, hence the redundancy. Second, you could have said that the "identity" of the group is $0$.
– Frenzy Li
Jul 15 at 12:06




1




1




You are correct in saying that the answer to your question is no.
– Derek Holt
Jul 15 at 12:42




You are correct in saying that the answer to your question is no.
– Derek Holt
Jul 15 at 12:42










2 Answers
2






active

oldest

votes

















up vote
1
down vote













When working in modulo $6$, notice that $0equiv 6bmod 6$; so actually your set in question is $0,1,2,3,4,5$.



Also note that the inverse of the group isn't $0$ - it is actually the identity element. To distinguish the difference between the two, recall the definitions



  • The identity element of a group $G$, $e$ say, is an element such that $acirc e=ecirc a=a$.

  • The inverse of an element $a$ in a group $G$ is an element $b$ such that $acirc b=bcirc a=e$ where $e$ is the identity element.

With this information in mind - now if you check the group axioms, you will find that this is indeed a group.






share|cite|improve this answer





















  • I think this is missing the point. You are not answering the question that was asked, which was "is $0,1,2,3,4,5,6$ a group under addition modulo $6$" and the answer to that is no. I am guessing that this was some kind of trick question, and you were supposed to answer the question that was asked, and not change the question to something different.
    – Derek Holt
    Jul 15 at 14:47


















up vote
-1
down vote













It is indeed a group. Associativity can easily be proven. The neutral element is 0 and each element has an inverse element, which you can see in the table (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image31.gif). If you want to check if it's a group you need each entry once per row/column in the table (and you need to prove associativity which can't be seen in the table). For example multiplication mod 6 isn't a group which you can see here (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image30.gif). Your group is also an abelian group, you can see commutativity in the table because it's symmetric.






share|cite|improve this answer

















  • 1




    It is not a group - to get a group you would need to remove one of the elements $0$ and $6$.
    – Derek Holt
    Jul 15 at 12:41











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2 Answers
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active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













When working in modulo $6$, notice that $0equiv 6bmod 6$; so actually your set in question is $0,1,2,3,4,5$.



Also note that the inverse of the group isn't $0$ - it is actually the identity element. To distinguish the difference between the two, recall the definitions



  • The identity element of a group $G$, $e$ say, is an element such that $acirc e=ecirc a=a$.

  • The inverse of an element $a$ in a group $G$ is an element $b$ such that $acirc b=bcirc a=e$ where $e$ is the identity element.

With this information in mind - now if you check the group axioms, you will find that this is indeed a group.






share|cite|improve this answer





















  • I think this is missing the point. You are not answering the question that was asked, which was "is $0,1,2,3,4,5,6$ a group under addition modulo $6$" and the answer to that is no. I am guessing that this was some kind of trick question, and you were supposed to answer the question that was asked, and not change the question to something different.
    – Derek Holt
    Jul 15 at 14:47















up vote
1
down vote













When working in modulo $6$, notice that $0equiv 6bmod 6$; so actually your set in question is $0,1,2,3,4,5$.



Also note that the inverse of the group isn't $0$ - it is actually the identity element. To distinguish the difference between the two, recall the definitions



  • The identity element of a group $G$, $e$ say, is an element such that $acirc e=ecirc a=a$.

  • The inverse of an element $a$ in a group $G$ is an element $b$ such that $acirc b=bcirc a=e$ where $e$ is the identity element.

With this information in mind - now if you check the group axioms, you will find that this is indeed a group.






share|cite|improve this answer





















  • I think this is missing the point. You are not answering the question that was asked, which was "is $0,1,2,3,4,5,6$ a group under addition modulo $6$" and the answer to that is no. I am guessing that this was some kind of trick question, and you were supposed to answer the question that was asked, and not change the question to something different.
    – Derek Holt
    Jul 15 at 14:47













up vote
1
down vote










up vote
1
down vote









When working in modulo $6$, notice that $0equiv 6bmod 6$; so actually your set in question is $0,1,2,3,4,5$.



Also note that the inverse of the group isn't $0$ - it is actually the identity element. To distinguish the difference between the two, recall the definitions



  • The identity element of a group $G$, $e$ say, is an element such that $acirc e=ecirc a=a$.

  • The inverse of an element $a$ in a group $G$ is an element $b$ such that $acirc b=bcirc a=e$ where $e$ is the identity element.

With this information in mind - now if you check the group axioms, you will find that this is indeed a group.






share|cite|improve this answer













When working in modulo $6$, notice that $0equiv 6bmod 6$; so actually your set in question is $0,1,2,3,4,5$.



Also note that the inverse of the group isn't $0$ - it is actually the identity element. To distinguish the difference between the two, recall the definitions



  • The identity element of a group $G$, $e$ say, is an element such that $acirc e=ecirc a=a$.

  • The inverse of an element $a$ in a group $G$ is an element $b$ such that $acirc b=bcirc a=e$ where $e$ is the identity element.

With this information in mind - now if you check the group axioms, you will find that this is indeed a group.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 15 at 11:57









thesmallprint

2,2191617




2,2191617











  • I think this is missing the point. You are not answering the question that was asked, which was "is $0,1,2,3,4,5,6$ a group under addition modulo $6$" and the answer to that is no. I am guessing that this was some kind of trick question, and you were supposed to answer the question that was asked, and not change the question to something different.
    – Derek Holt
    Jul 15 at 14:47

















  • I think this is missing the point. You are not answering the question that was asked, which was "is $0,1,2,3,4,5,6$ a group under addition modulo $6$" and the answer to that is no. I am guessing that this was some kind of trick question, and you were supposed to answer the question that was asked, and not change the question to something different.
    – Derek Holt
    Jul 15 at 14:47
















I think this is missing the point. You are not answering the question that was asked, which was "is $0,1,2,3,4,5,6$ a group under addition modulo $6$" and the answer to that is no. I am guessing that this was some kind of trick question, and you were supposed to answer the question that was asked, and not change the question to something different.
– Derek Holt
Jul 15 at 14:47





I think this is missing the point. You are not answering the question that was asked, which was "is $0,1,2,3,4,5,6$ a group under addition modulo $6$" and the answer to that is no. I am guessing that this was some kind of trick question, and you were supposed to answer the question that was asked, and not change the question to something different.
– Derek Holt
Jul 15 at 14:47











up vote
-1
down vote













It is indeed a group. Associativity can easily be proven. The neutral element is 0 and each element has an inverse element, which you can see in the table (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image31.gif). If you want to check if it's a group you need each entry once per row/column in the table (and you need to prove associativity which can't be seen in the table). For example multiplication mod 6 isn't a group which you can see here (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image30.gif). Your group is also an abelian group, you can see commutativity in the table because it's symmetric.






share|cite|improve this answer

















  • 1




    It is not a group - to get a group you would need to remove one of the elements $0$ and $6$.
    – Derek Holt
    Jul 15 at 12:41















up vote
-1
down vote













It is indeed a group. Associativity can easily be proven. The neutral element is 0 and each element has an inverse element, which you can see in the table (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image31.gif). If you want to check if it's a group you need each entry once per row/column in the table (and you need to prove associativity which can't be seen in the table). For example multiplication mod 6 isn't a group which you can see here (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image30.gif). Your group is also an abelian group, you can see commutativity in the table because it's symmetric.






share|cite|improve this answer

















  • 1




    It is not a group - to get a group you would need to remove one of the elements $0$ and $6$.
    – Derek Holt
    Jul 15 at 12:41













up vote
-1
down vote










up vote
-1
down vote









It is indeed a group. Associativity can easily be proven. The neutral element is 0 and each element has an inverse element, which you can see in the table (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image31.gif). If you want to check if it's a group you need each entry once per row/column in the table (and you need to prove associativity which can't be seen in the table). For example multiplication mod 6 isn't a group which you can see here (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image30.gif). Your group is also an abelian group, you can see commutativity in the table because it's symmetric.






share|cite|improve this answer













It is indeed a group. Associativity can easily be proven. The neutral element is 0 and each element has an inverse element, which you can see in the table (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image31.gif). If you want to check if it's a group you need each entry once per row/column in the table (and you need to prove associativity which can't be seen in the table). For example multiplication mod 6 isn't a group which you can see here (http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/Images/image30.gif). Your group is also an abelian group, you can see commutativity in the table because it's symmetric.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 15 at 12:04









Vajk

283




283







  • 1




    It is not a group - to get a group you would need to remove one of the elements $0$ and $6$.
    – Derek Holt
    Jul 15 at 12:41













  • 1




    It is not a group - to get a group you would need to remove one of the elements $0$ and $6$.
    – Derek Holt
    Jul 15 at 12:41








1




1




It is not a group - to get a group you would need to remove one of the elements $0$ and $6$.
– Derek Holt
Jul 15 at 12:41





It is not a group - to get a group you would need to remove one of the elements $0$ and $6$.
– Derek Holt
Jul 15 at 12:41













 

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