Symmetric difference of cycles Let C and C' be cycles in a graph G. Prove that the symmetric difference of C and C' decomposes into cycles.
I know that the symmetric difference of C and C' contains only the edges that are in exactly one of C or C'. Since both C and C' are cycles, so d(v) for all v in V(G) must be even. So the difference would include a bunch of edges whose endpoints have even degree. If the degree is at least 2, we have cycle(s). But what if degree = 0? I am not really sure if it could be 0. I need a leg up, please! Thanks.
 A: In the context of a question like this one you are only interested in edges
and it is convenient to 'disregard' vertices.
After all you want the symmetric difference of two cycles that have exactly one vertex in common
to be just the same two cycles: you do not want to 'subtract a vertex'.
You also do not want to answer the question: how many vertices has the symmetric
difference of a cycle with itself.
So we will simply assume that you have some underlying set of vertices,
some of which are spanning your cycles, but we do not care about the vertices.
This answers your question "But what if degree=0?": if the degree is 0, there
are no incident edges, so we do not care.
For the rest you can follow the argument you already started:
first show that all degrees in the symmetric difference of two cycles are even,
then show that any graph with only even degrees decomposes into cycles.
Let me know if you need help with any of these parts.
Note that your proof will in fact demonstrate a stronger result:
you can have an arbitrary number of cycles, but as long as you only use
the operations union and symmetric difference the result can be decomposed into cycles.
