The Hamiltonian cycle problem is P-hard?

I know what P-hard means. Let's denote the P-hard problem as H. For any problem A in P, there exists a polynomial-time reduction from A to H. I think the answer is yes (I suppose that every "easy" problem can be reduced to a "harder" problem (in this case problems in P to an NP-complete problem), but I am not sure.

• What is the question? Make sure to include the question in the body of the post. Please check the definition of P-complete and P-hard -- does it refer to polynomial-time reductions? I suggest you include the definition of P-hard that you are using. "I suspect the answer..." - what answer? answer to what question? I notice you've received similar feedback before: cs.stackexchange.com/questions/165166/…
– D.W.
Jan 23 at 10:40

The Hamiltonian cycle problem is $$\mathsf{NP}$$-Hard (w.r.t. poly-time reductions), that means that there is a Karp reduction from any problem in $$\mathsf{NP}$$ to Hamiltonian cycle.
Since $$\mathsf{P} \subseteq \mathsf{NP}$$ the same holds true for any problem in $$\mathsf{P}$$. Therefore Hamiltonian cycle is $$\mathsf{P}$$-Hard (w.r.t. poly-time reductions).
Note however, that stronger notions of reductions are often used when talking about $$\mathsf{P}$$-completeness / $$\mathsf{P}$$-hardness. Using poly-time reduction yields some "rough" results like: All non-trivial problems in $$\mathsf{P}$$ are $$\mathsf{P}$$-Hard.