group-theory
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?
The video states that "the set of all cosets forms a group". This is exactly the set $\mathbb{Z}/n\mathbb{Z}$ as you know it. It doesn't say that a coset itself is a subgroup of $\mathbb{Z}$. And indeed, the coset $3 + 5\mathbb{Z}$ is not a subgroup of $\mathbb{Z}$, as you noted. The set $3 + 5\mathbb{Z}$ could be turned into a group though, if we define some silly operation on it that has the identity element $-12$ for example.
I think all you need has been pointed here. You don't need to write down the associated Cayley table for the presented set and its operation, but for this one it leads you to get the answer graphically:
We can see that the operation is a binary one. Can you find the identity element? What about the inverses ones?
Best Answer
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
With this information in mind - now if you check the group axioms, you will find that this is indeed a group.