Clear Definition of Two Graphs Being Isomorphic I am wanting to deal with canonical graph homorphisms and other graph homorphisms much more deeply, but I need a definition for isomorphism between graphs to start. I know the definition of two groups being isomorphic, but I do not know how to tell if two graphs are isomorphic to each other. I watched a video which explains how to tell if two graphs are isomorphic. Now, the video states the following definition of isomorphic graphs: $G_1\cong G_2$ iff $\exists f : V(G_1)\xrightarrow[]{\text{bijection}} V(G_2)$ ST $f(u)f(v)\in G_2 \leftrightarrow uv\in G_1$. Personally, I find this definition hard to work with for two reasons: 


*

*Reason 1: How can I draw a table similar to that of groups in abstract math?

*Reason 2: How is there even an operation defined in this way? It seems as though the operation is not closed (it also seems to be more of a directed graph than an undirected graph in the way that it is not a set), and it does not resemble that of the abstract algebra definition which would be in this case by the functions codomain the following: $G_1\cong G_2$ iff $\exists f : V(G_1)\xrightarrow[]{\text{bijection}} V(G_2)$ ST $\forall u, v \in V(G_1)$, [$f(u\cdot v)=f(u)\diamond f(v)$].





*

*QUESTION:

Is there a good definition of isomorphism between graphs that I should be using by you graph theory/abstract algebra/other experts? Any help would be greatly appreciated!


Above are two graphs which I represented as a table like that of an adjacency matrix if this helps answer the question by giving an example to talk about. So,  $G_1:=(V(G_1), E(G_1))$ (shown on the left) and $G_2:=(V(G_2), E(G_2))$ (shown on the right). 
So, in this context, $f=\begin{pmatrix}
v_1 & v_2  & v_3 \\ 
v_4 & v_5 & v_6
\end{pmatrix}$.
\begin{array}{c||c||c||c}
Adjacency(G_1) & v_1 & v_2 & v_3 \\
\hline
v_1 & 0 & 2 & 0  \\
\hline
v_2 & 2 & 0 & 1  \\ 
\hline
v_3 & 0 & 1 & 0  
\end{array}
\begin{array}{c||c||c||c}
Adjacency(G_2) & v_4 & v_5 & v_6 \\
\hline
v_4 & 0 & 2 & 0  \\
\hline
v_5 & 2 & 0 & 1  \\ 
\hline
v_6 & 0 & 1 & 0  
\end{array}

Consider the following definition of group isomorphism: it is a bijection $f:G_1\to G_2$ such that
$$uvw^{-1}=e_{G_1} \iff f(u)f(v)f(w)^{-1}=e_{G_2}\quad \forall u,v,w\in G_1.$$
And now compare graphs:
$$uv\in E(G_1) \iff f(u)f(v)\in E(G_2)\quad \forall u,v\in V(G_1).$$
Not much difference really.
