Is the number of simple circuits of a particular length preserved in two isomorphic graphs?

Question

If two graphs are isomorphic, and one has a simple circuit of a particular length, must the other graph also have a circuit of the same length?

Do they also have to have the same number of such circuits?

In the book that I use it is claimed that the following two graphs are isomorphic:

Graph G has two circuits of length 5, but no circuits of length 4. Graph H has one circuit of length 5 and one circuit of length 4. Doesn't this mean that they are not isomorphic?

Edit: The book shows their isomorphism with the function f, where f(u1)=v6, f(u2)=v3, f(u3)=v4, f(u4)=v5, f(u5)=v1, f(u6)=v2. Doesn't this prove that they are isomorphic?

Answer 1

Yes, it does mean that they’re not isomorphic. If they were isomorphic, an isomorphism $h:V(H)\to V(G)$ would carry the $4$-cycle with vertices $v_3,v_4,v_5$, and $v_6$ to a $4$-cycle in $G$. Since $G$ has no $4$-cycle, no such isomorphism can exist.

Answer 2

Unless the question has been edited and changed sometime in the last two years, the book is correct and the graphs are isomorphic.

It is true that isomorphic graphs must have the same number of circuits of any given length. However, you are incorrect when you say that this is false in the example you have given.

$G$ does have a circuit of length $4$, it is the "outside" rectangle consisting of $u_1,u_3,u_4,u_5$ (although there is a problem with your labelling, you have two vertices called $u_5$).

$H$ has two circuits of length $5$, namely, $v_1,v_2,v_3,v_6,v_5$ and $v_1,v_2,v_3,v_4,v_5$.