The question that I'm trying to answer is: "Show that both graphs drawn below are isomorphic by giving an isomorphism for them. Also known as the Petersen Graphs.
I dont quite understand what is meant by 'give an isomorphism'. What is the best method to answer questions such as this one?
On
Knowing nothing about the Petersen graph, I decided to follow Lord Shark's advice:
"Just pair off the vertices in one with the vertices in the other, so that adjacency is preserved."
The hard part would be guessing how to pair off the vertices; once that was done, it would be easy to check if adjacencies were preserved.
The first thing I noticed about the graph on the right is that it has a cycle of length nine: $1,2,3,6,8,10,9,7,4.$ If the graphs are isomorphic I should be able to find a nine-cycle in the other one. Sure enough, after a couple of false starts, I found one: $a,f,k,i,g,d,h,c,b.$ After that I just did what came naturally and tried matching one cycle with the other: $1\to a,\ 2\to f,\ 3\to k,\ 6\to i,\ 8\to g,\ 10\to d,\ 9\to h,\ 7\to c,\ 4\to b$ and of course paired the leftover vertices: $5\to e.$
Then I check adjacencies. Luckily, everything worked!
A sample isomorphism for your graphs is relatively simple here if you know that the Petersen graph is strongly regular, meaning that the choice of starting point for your isomorphism is arbitrary.
So we can map between the two diagrams starting with an arbitrary three-point arc, so taking mapping pairs of $(a,1),(b,2),(c,3)$ we can continue the 5-cycle with $(e,9),(f,10)$ then completing adjacent sets we have $(d,4), (i,5), (k,8), (h,6),(g,7)$ giving a full isomorphism:
$$(a,1),(b,2),(c,3),(d,4),(e,9),(f,10),(g,7), (h,6), (i,5), (k,8)$$
Due to the strong regularity, we can pick any starting point, any of its adjacent points then any further adjacent point to map to a given arc of three points on one graph, giving $10\times 3\times 2= 60$ different isomorphisms to choose from.