# If $K$ is a subgroup of $G$, then $\phi(K) = \{ \phi(k) | k \in K \}$ is a subgroup of $\bar{G}.$ Vice-versa?

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.

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Just because they don't prove that the statement is "if and only if" doesn't mean that it isn't; in this case, as your argument shows, it definitely is. I suspect the omission was either unintentional, or because the author thought the "only if" part obvious and wanted to leave it to the reader to figure out for themselves. – Alex Becker Jan 18 '12 at 23:41
possible duplicate of Image of subgroup and Kernel of homomorphism form subgroups – François Muer Jan 8 '14 at 13:28

This is an if and only if statement, under the assumption that $\phi$ is an isomorphism. However, a generalization of the statement you made is often given as "If $\phi:G\to\bar G$ is a homomorphism and $K$ is a subgroup of $G$, then $\phi(K)$ is a subgroup of $\bar G$". This is what I believe the reference was referring to, and it is not an if and only if statement (we can consider for example the trivial isomorphism, which sends any subset of $G$ to a subgroup of the trivial group). Note that if $\phi$ is an isomorphism, then $\phi,\phi^{-1}$ are isomorphisms so this becomes an if and only if statement.