Solving subgraph isomorphism in polynomial time
Here is a more general theorem:


  Let $X$ be some NP-complete problem. Then $X$ can be solved in polynomial time iff all problems in NP can be solved in polynomial time.


Indeed, if all problems in NP can be solved in polynomial time, then since $X$ is in NP, then $X$ can be solved in polynomial time.

In the other direction, suppose that $X$ can be solved in polynomial time. Let $Y$ be any problem in NP. Since $X$ is NP-hard, there is a polytime reduction $f$ from $Y$ to $X$. Using this reduction, we can reduce $Y$ to $X$, and so solve $Y$ in polynomial time. (I'm only sketching this part, since this is a standard argument which I assume you were shown in class.)