Suppose $\phi: G \to \bar{G}$ is an isomorphism from one group to the other. Then the following is true: If $K$ is a subgroup of $G$, then $\phi(K) = \{ \phi(k) | k \in K \}$ is a subgroup of $\bar{G}.$ However, my reference makes a point of emphasizing that this is not an if and only if statement. I struggle to understand why this wouldn't work in the other direction. If $\phi(K) = \{ \phi(k) | k \in K \}$ is a subgroup of $\bar{G},$ wouldn't $\phi^{-1} (\phi(K))$ produce a subgroup of $G,$ since $\phi^{-1}$ is an isomorphism (so one-to-one, onto, preserves the operations)? Or do I not understand the "other direction"/"and only if" part correctly?
Admittedly, the book "doesn't make a point" in a sense of actually stating it in words, but I think it does in a sense that the previous three properties are listed as if and only if, and this one is just "if," which seemed an important enough difference for a mathematics book.