Prove that if G is bipartite, then there does not exists a subgraph g in G such that g is a cycle with odd number of edges?

Question

I understand the intuition for this lemma but having a hard time proving it formally. I was thinking of using induction to show P(n)" graph with even edges" implies P(k+1)"an acyclic subgraph with odd edges". But what type of graph would you end up getting if you remove 1 edge from p(k+1), there are many types of even lengthed subgraphs,specifically that even lengthed subgraph is indeed a cycle or it is not? Thanks!

Answer 1

Assume the sequence of vertices $v_0,v_1,\ldots, v_n$ represents a cycle (so $v_0 = v_n$). Inductively show that $v_0$ and $v_{2k+1}$ are in different parts for every $k \geq 0,$ so $v_0 = v_n$ implies $n$ is even.