I know that for $n \ge 3$, $S_n$ is non-Abelian. I would like to work out an example in terms of graphs so to make it sure that I understand it right.
A symmetric group of graphs of four vertices, $G$, could be considered as the set of all of their adjacency lists. The group operation is composition of permutations on graph vertices. The order of the group is $4! = 24$.
Let, $g$ be a graph as follows,
So, a group operation could be $(0 1) \circ (1 3)$ which renders the following graph which is also an element of $G$.
I am confused about the identity element. I understand that the trivial permutation keeps the graph the same. But as the elements of the group are the graphs of four vertices, what is the identity element in the form of a graph?
Moreover, I have no clue what is the inverse here.
I think the non-Abelianness is obvious. If we change the previous composition, i.e., $(1 3) \circ (0 1)$ renders a different graph which is as follows,
Is it correct what I have understood so far? Can anyone help me understanding the inverse and identity elements? Thanks!
It's not clear what you mean by "non-Abelianness of symmetric group for graphs," but I'll comment on your example. The symmetric group $S_n$ does act on the set of labeled graphs on $n$ vertices in the manner you describe. The automorphism group of a graph is sometimes referred to as the "symmetry group of the graph" or simply "the group of the graph".
You started with a graph on vertex set $V=\{0,1,2,3\}$ and edge set $\{01, 12, 23, 03, 02\}$, and obtained other graphs on the same vertex set by relabeling the vertices according to the action of certain (compositions of) transpositions. The symmetric group on $V$ (which is isomorphic to $S_4$) acts on the 2-subsets of $V$ and hence on the edge set, in the manner you describe (i.e. by relabeling the points). Let $V^{(2)}$ denote the set of all 2-subsets of $V$, and let $\mathcal{G}$ denote the set of all subsets of $V^{(2)}$. Thus, if $|V|=n$, then $|V^{(2)}|={n \choose 2}$, and $|\mathcal{G}|=2^{{n \choose 2}}$. Then each labeled graph on $V$ (i.e. each element of $\mathcal{G}$) corresponds to a unique subset of $V^{(2)}$ (the correspondence is that if the graph has edge-set $E$, then $E$ is this unique subset).
Starting with your given graph, by continuing your relabeling process using all transpositions (or all permutations) in $Sym(V)$, you would obtain all the edge-sets (and hence all labeled graphs in $\mathcal{G}$) that lie in the orbit of the given graph in the action of $Sym(V)$ on $\mathcal{G}$. In your case, you would obtain a total of 6 distinct labeled graphs. (One way to obtain this number is to just observe that the number of distinct graphs is the number of ways to pick two distinct vertices from the four vertices for the vertices of degree 3, and this can be done in ${4 \choose 2}=6$ ways. Another way to prove this is using the orbit-stabilizer lemma. $Sym(V)$ acts on $\mathcal{G}$, and the stabilizer subgroup that fixes the given graph is its automorphism group, which is the Klein group $C_2 \times C_2$ of order 4. By the orbit-stabilizer lemma, the size of the orbit is $4! / 4 = 6$.)
The image of the action of $Sym(V)$ on $\mathcal{G}$ is a subgroup of $Sym(\mathcal{G})$. Your example shows (and this is probably what you meant) that this image is a nonabelian subgroup. Actually, we can restrict the action of $Sym(V)$ to the subset $\mathcal{H} \subseteq \mathcal{G}$ containing just the six graphs mentioned in the previous paragraph. You essentially showed that the image of this restricted action, which is a permutation group in $Sym(\mathcal{H})$, is nonabelian.
While $Sym(V) \cong S_4$ is not abelian, the graph you've drawn has automorphism group $C_2 \times C_2$, which is abelian. If you are looking for a graph whose automorphism group is nonabelian, just take the complete graph on 3 vertices. It's automorphism group is $S_3$, which is the smallest nonabelian group (all groups of order at most 5 are abelian).