there is an example that my lecturer went through which is to prove (Z3,+)is a group. He prove this using a cayley table. I am stuck at the part when he mentioned:
[0]+[2]=2
[1]+[2]=[3]=[0]
and also when he prove that it satisfy the inverse, he wrote down[0]-1=[0] and [1]-1=[2]. I am not to sure how he get this result from the table. Thank you!
The Cayley table of $(\mathbb{Z}/3\mathbb{Z}, +)$ is
\begin{array}{c|c|c|c|c} + & [0] & [1] & [2]\\ \hline [0] & [0] & [1] & [2] \\ \hline [1] & [1] & [2] & [0]\\ \hline [2] & [2] & [0] & [1]\\ \end{array}
and recall that within this group, the binary operation between elements is defined as: $$(\forall a,b \in \mathbb{Z}/3\mathbb{Z}): \qquad a+b := (a+b) \text{ mod }3$$ which explains why in the table we see things like $[2]+[2]$, which traditionally you would imagine is $[4]$, but in this case we have $4 \text{ mod }3 \equiv 1$, hence $[2]+[2]=[1]$.
Going a bit deeper assuming you know what an equivalent class is, the reason we've been putting brackets around everything is because additional modulo 3 is an equivalence relation and things like $[4]=[1]$ and $[234]=[0]$ are valid in this group because they're equivalent, that is, of the same class. An important thing to know is that addition in this group is a well-defined operation, meaning that since we have $[4]=[1]$ and $[234]=[0]$ (that is, $4 \equiv 1 \text{ mod } 3$ and $234 \equiv 0 \text{ mod } 3$) that the following should be true: $$[4]+[234]=[1]+[0].$$