I was requested to show the following graphs are not isomorphic. I started studying graph theory literally half an hour ago, and I'm supposed to be able to show this without any knowledge of degrees and paths (since this problem comes immediately after defifining what a graph is, what an isomorphism is, and what a subgraph is).
I had to use a totally intuitive conception of paths to find a way to prove this. Again, I have zero formal knowledge on paths so my notation, terminology, etc., might be quite off. However, I'm mostly concerned with whether I'm understanding the core ideas of subgraphs and isomorphisms right. Here's the answer as I wrote it.
$\text{Answer.}$
Observation. There exists at least a subgraph $G'_1$ of $G_1$ with a $3-$element subset $V_1'$ of $V_1$ such that $v'_i$ exists in two different edges $e_k, e_j \in E'_1$.
Note. This is my brute way of saying that in $G_1$ one can start at a point $v_i$ with the possibility of returning to it following a path of $3$ vertices.
No such subgraph $G'_2 \subset G_2$ satisfies this same property.
Trying to mean: it is impossible to return to any initial $v_i \in V_2$ through a path of $3$ vertices (one must take at least a $4$-vertices path).
Answer. We know (I skip the theorem) any isomorphism $\alpha$ between two graphs will map a subgraph to an isomorphic subgraph. However, the previous observation shows there are $3-$element subgraphs of $G_1$ that can not be appropriately mapped to an isomorphic subgraph of $G_2$. Then the graphs are not isomorphic.
$\text{Side note}$.
I alternatively considered to state that there are subgraphs of $3$ elements of $G_1$ that have $3$ edges, with no subgraphs of $3$ elements of $G_2$ satisfying that property. But it seemed like I was saying the same thing in different words.
I'm not only concerend about whether the statements themselves are correct or not, but also about whether even being true they are enough to prove what is requested. Thanks in advance.
Best Answer
Your argument is correct and quite enough to prove what you claim. A shorter way to say it is to note that the first graph contains triangles and the second does not.
"Having triangles" is a property clearly preserved by graph isomorphism.
Relying on your intuitive grasp of paths even before encountering a formal definition is fine. It's actually better than fine, since it will make it easier to understand the formal definition when you see it.