Graph automorphisms and the diamond graph

Question

I thought I understood graph automorphisms, but looking at a specific example is tripping me up. Consider the diamond graph: Apparently, the Klein 4-group is the automorphism group of this graph. The Klein 4-group is $$\{(), (12)(34), (13)(24), (14)(23)\}$$

So, if we take $\pi=(12)(34)$ as an example, the definition of graph isomorphism says that $ij$ is an edge in $G$ if and only if $\pi(i)\pi(j)$ is an edge in $G$. But if we look at nodes $2,4$, then $24$ is an edge in $G$ but $\pi(2)\pi(4) = 13$ is not an edge in $G$. So why is $(12)(34)$ an automorphism for this graph?

What am I doing wrong? What is the best way to visualize the action of a permutation of the vertices of a graph?

Answer 1

The Klein $4$-group is the name given to any group with $4$ elements with each element being self-inverse. The graph automorphism group of a graph on $4$ vertices is, as you wrote, a subgroup of the symmetric group on $4$ elements. The graph automorphism group of the diamond graph you drew is $$ \{(), (13), (24), (13)(24)\} $$

This group is a Klein $4$-group, just not the one that you wrote.

Answer 2

Hint: Consider $\{(), (24), (13), (13)(24)\}$.