How are these graphs isomorphic?

Question

When I look at the first graph (graph G) there are 2 pentagons (2 circuits that have 5 vertices). The first one is $u_1,u_3,u_4,u_7,u_8$ and the second one $u_1,u_2,u_5,u_6,u_8$. In the second graph (graph H) there are 4 circuits that have 5 vertices. First $v_2,v_3,v_4,v_5,v_6$, second $v_2,v_1,v_8,v_7,v_6$, third $v_2,v_3,v_4,v_8,v_1$ and fourth $v_6,v_5,v_4,v_8,v_7$. So how are these graphs isomorphic? Maybe one of you sees other evidence that they are not isomorphic. Please help.

(This is a question from the Rosen's Discrete Mathematics and It's Applications Book - Chapter 10.4 - Problem 22)

Answer 1

You forgot two pentagons in graph G, which are harder to spot: $u_1, u_3, u_4, u_5, u_2$ and $u_4, u_5, u_6, u_8, u_7$.

To construct an isomorphism, you need to see that the 3-cycles are connected to each other with a single edge (from $u_1$ to $u_8$, and from $v_2$ to $v_6$). A possible solution would be:

• $u_1 \to v_2$
• $u_2 \to v_1$
• $u_3 \to v_3$
• $u_4 \to v_4$
• $u_5 \to v_8$
• $u_6 \to v_7$
• $u_7 \to v_5$
• $u_8 \to v_6$