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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 $6+_{6}0=0$ and $6+_{6}6=0$, 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?

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    $\begingroup$ 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. $\endgroup$ – Arnaud Mortier Jul 15 '18 at 11:51
  • $\begingroup$ Didn't get you Arnaud :( $\endgroup$ – user3767495 Jul 15 '18 at 11:53
  • $\begingroup$ @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 $0$. $\endgroup$ – Frenzy Li Jul 15 '18 at 12:06
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    $\begingroup$ You are correct in saying that the answer to your question is no. $\endgroup$ – Derek Holt Jul 15 '18 at 12:42
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When working in modulo $6$, notice that $0\equiv 6\bmod 6$; 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 $a\circ e=e\circ a=a$.
  • The inverse of an element $a$ in a group $G$ is an element $b$ such that $a\circ b=b\circ a=e$ 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.

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  • $\begingroup$ 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. $\endgroup$ – Derek Holt Jul 15 '18 at 14:47
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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.

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    $\begingroup$ It is not a group - to get a group you would need to remove one of the elements $0$ and $6$. $\endgroup$ – Derek Holt Jul 15 '18 at 12:41

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