I just start studying graph theory as found it's weird mathematics area I want to learn from math point view ,but when I try to find clip in youtube,mostly mention from com sci point view ok that is not problem in this question
This question ask about Isomorphism in Graph theory 
I just start read a few page of graph theory book I don't know to prove in graph theory term. But I borrow the term from group theory(even I don't know operation or the elenment that i define is group or not) define group G has order 3 (e,c,d) has operation "draw a line to each element" and do nothing is identity of this group and H as order 2 (C',D') as the "do notthing" each elemnt (in this case the i don't know how to define idenitiy)
since order is not equal it can't mapping bijective to each group therefore it's not isomorphism Can I use this similiar proof in isomorphism proof in graph theory ?
Isomorphism between graphs is "the same" as isomorphism between groups in the sense that both are isomorphisms in the sense of category theory:
In group theory, the objects are groups, i.e., tuples $\langle G,\circ\rangle$ of a set $G$ and a corresponding binary operation $\circ\colon G\times G\to G$ with certain properties. In graph theory, the objects are graphs, i.e., tuples $\langle V,E,I\rangle$ where $V$, $E$ are sets (of vertices and edges, respectively) and $I\subseteq V\times E$ is the incidence relation. Note that this is not absolute: Depending on what you consider a graph (directed? indirected? simple?), it may be the case that a graph is modelled in a different way, for example, for simple graphs we could define a graph simply as a tuple $\langle V,E\rangle$, where $V$ is a set and $E\subseteq\mathcal P(V)$ with $|e|=2$ for all $e\in E$.
In group theory, a (homo-)morphism from $\langle G,\circ\rangle $ to $\langle G',\circ'\rangle $ is a map $f\colon G\to G'$ such that $f(a\circ b)=f(a)\circ'f(b)$ for all $a,b\in G$. In graph theory, a morphism from $\langle V,E,I\rangle $ to $\langle V',E',I'\rangle $ is a pair $(f_V,f_E)$ of maps $f_V\colon V\to V'$, $f_E\colon E\to E'$ such that $\langle f_V(v),f_E(e)\rangle\in I'$ whenever $\langle v,e\rangle\in I$. (Or with the alternative described above, a morphism from $\langle V,E\rangle$ to $\langle V',E'\rangle$ is a map $f\colon V\to V'$ with $\{f(a),f(b)\}\in E'$ for all $\{a,b\}\in E$).
We have a composition of morphisms, i.e., given a morphism $\phi\colon A\to B$ and a morphism $\psi\colon B\to C$, we define a suitable morphism $\psi\circ \phi\colon A\to C$. In both group and graph theory, the composition of morphisms is obtained by composing the underlying maps. As required in category theory, the composition thus defined is associative (because the composition of the underlying maps is associative).
For each object, we always have an identity morphism, i.e., a morphism that behaves neutral with respect to composition: in group theory from the identity map $\operatorname{id}_G\colon G\to G$, in graph theory from the pair $\langle \operatorname{id}_V,\operatorname{id}_E\rangle$ of identity maps of $V$ and $E$.
As in every category, an isomorphism from object $A$ to object $B$ is defined as a morphism $\phi\colon A\to B$ such that there exists a morphism $\psi\colon B\to A$ such that $\psi\circ \phi=\operatorname{id}_A$ and $\phi\circ \psi=\operatorname{id}_B$.
Breaking down these abstract remarks, we find that an isomorphimsm between two graphs is a bijective map between the vertex sets together with a compatible bijection between the edge sets (or: a bijection between the vertex sets such that the image vertices a joined by an edge oif and only if the original vertices are).
So, apart from the common underlying framework o fcategory theory, there is not really a connection that allows us e.g. to identify edges with group elements in as simple a way as you try in your paragraph (nevertheless, there are ways to represent groups as specific graphs).
In the concrete example, the simplest way to show that the graphs are not isomorphic is by exhibiting properties that are left invariant by an isomorphism, but are different in the two graphs. For example, the number of vertices could such an invariant, however, both graphs have $8$ vertices, so theymight still be isomorphic. They also bot have the same bnumber of edes, the same number of degree 2 vertices, and moch more. But: In the left graph there is no edge with bot ends having degree $2$, whereas the graph on the right has two such edges. We conclude that the graphs are not isomorphic (be sure to understand why "the number of edges with both end having degree $2$" is an invariant!)