Group under Additive modulo 6 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?
 A: 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.
A: 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.
