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Let $G=(V,E)$ be a simple undirected graph, where $V$ is the vertex set and $E$ the edge set. Let $|V|\leq|E|$. I need to prove that $G$ has a simple cycle. My approach was - assume $G$ has no simple cycles. If it's connected, that it's a tree, and then $|E|=|V|-1$, and that contradicts the assumption $|V|\leq|E|$. However I can't find any contradiction to the assumption if G has no simple cycles and isn't connected. Can you give me a hint?

Note: I'm only just beginning graph theory - we haven't learn Hamiltonian or Eulerian graphs, or isomorphisms, or loops. All we know are degrees, paths, cycles, trees, subgraphs and components and several simple theorems about them.

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If $G$ is not connected, but has no cycle, its connected components are trees ($G$ is called a forest then). That is, for each component $C$ of $G$ we have $$ |V(C)| = |E(C)| + 1 $$ Adding over all compontents, we get, if $G$ has $k$ components, $$ |V(G)| = |E(G)| + k $$

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    did you mean |V(C)| -1 = |E(C)|? – 2017-01-06
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    Sure, thanks, corrected. – 2017-01-06
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  1. If $G$ isn't connected, it can be split into components, $G_1,\dots, G_k$.
  2. If all the components are trees, then $|V|\leq |E|$ is not possible
  3. Therefore, one of the connected components is not a tree, and has a cycle.