How to show that two groups are isomorphic? Consider the groups $G = \{0,1,2\} = \mathbb Z_3$ and $H = \{a,b,c\}$
given by the following multiplication tables:

The first one isn't really multiplication but in my notes it said it doesn't really matter.
So how do I show an isomorphism? The groups have the same size so they can be bijective right? But it just seems so abstract to show if there's an isomorphism... Exactly what do we have to check.
 A: First, find the identity in each. In your first example, it’s “$0$”, while in the second it’s “$b$”. How to see this? It’s the element whose row and column matches the labeling rows. Now see whether you can match some nonidentity element in the first example to a nonidentity element in the second so that you force a homomorphism.
A: First check the second table indeed defines a group law. Looking at the second row and second column in the right-hand side table, you can see $b$ is the neutral element. Then you can check $a^{-1}=b$  and $b^{-1}=a$. You can at once see that the law is commutative (symmetry w.r.t. the principal diagonal). Also, which is longer, but shortened by commutativity, you should check the law is associative, examining all possibilities for the triples $(a,b,c)$.
Now you can define a bijective mapping from the first set to the second by
\begin{align*}
0&\mapsto b,\\
1&\mapsto a\quad (\text{or}\enspace b),\\
2&\mapsto c\quad (\text{or}\enspace a).
\end{align*}
There remains to check it is compatible with both laws, considering all possible cases.
A: Suppose that there is an isomorphism $\phi$.
Then you must have $\phi(0 + 0) = \phi(0) * \phi(0)$, so
from the table we must have $\phi(0) = b$.
Now there are only two possibilities for $\phi(1),\phi(2)$.
If we try $\phi(1) = a,\phi(2) = c$, it is easy to check that
$\phi$ is an isomorphism (that is, $\phi(x+y) = \phi(x) * \phi(y)$ and $\phi$ is a bijection).
In fact, if we let $\phi(1) = c,\phi(2) = a$, it is easy to check that
$\phi$ is also an isomorphism.
A: $O$ is the identity in G and $b$ is the identity in H. Another thing to notice is that the tables are symmetric. Hence $G,H$ are commutative groups.
Take a look at $\psi:H\rightarrow G$
$\psi(b)=0$
$\psi(a)=1$
$\psi(c)=2$

$\psi(a*c)=\psi(b)=\psi(a)+\psi(c)...$
