Disjoint union of cycles 
Prove that every 2-regular graph is isomorphic to a disjoint Union of cycles.

What does a disjoint Union of cycles mean?
And how is it isomorphic to a 2-regular graph?
I'm new to graph theory, I understand what a 2-regular graph is and what isomorphism is. I don't understand how they connect since I don't understand what a disjoint Union of cycles would mean. 
 A: I will assume you encountered a problem saying "Prove every 2-regular graph is a disjoint union of cycles" and talk about that.
Suppose a finite graph is 2-regular. Pick a vertex. It has two edges. Follow along one of those edges to get another vertex. That vertex is connected by another edge to another vertex. And so on. In this way we get a sequence of vertices each determined by the previous one.
Eventually, since there are only finitely many vertices, this sequence of vertices must repeat itself, and so we've gone in a circle. In other words, we've found a cycle within the graph. And this cycle includes a pair of edges for every vertex, which means there are no other edges connecting these vertices to anything outside the cycle. In other words, this cycle is disjoint from the rest of the graph.
Then we can move this known cycle to the side and repeat the process over and over again with the rest of the graph until we've encountered every vertex. In this way, we see that the graph is a bunch of cycle graphs that are all disjoint from each other.
For example, here is a 2-regular graph with 15 vertices:
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