Let $G$ be a finitely generated group and let $\Gamma=Cay(G, S)$ be the Cayley graph of $G$ with respect to some generating set $S$.


If there exists $S$ such that $\Gamma$ is bipartite, then $G$ has an index $2$ subgroup (e.g. generated by all words of even length). Does the converse hold? (I'd guess not, but I don't know generic invariants allowing to show that a group does not admit a bipartite Cayley graph)



Does index 2 subgroup imply bipartite Cayley graph?
**Is** bipartite or **admits** bipartite? To the first question the answer is obviously "no", to the second, obviously "yes". Let $H\subset G$ be an index $2$ subgroup; take generators $h\_1,h\_2,\ldots$ for $H$ and one more generator $g\notin H$. In this generating set, the Cayley graph is not bipartite (with a few possible exceptions). However, if you switch to $g, h\_1g, h\_2g,\ldots$, this is also a generating set and all generators have "odd degree", so the graph is obviously bipartite. 


