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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 –  Frank Jan 8 at 13:28

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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.

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Thanks for your answer. I do not think it was referring to homomorphisms because we haven't studied them yet and they are not described in the chapter. Would posting a picture of the page or the pdf reference from google books etc. help in clarifying the context? I will look for one right now. –  questionado Jan 18 '12 at 23:20
    
A picture would definitely help. But I believe it was referring to homomorphisms because 1) its true about them and 2) the homomorphism statement is typically one of the first things you learn about homomorphisms. –  Alex Becker Jan 18 '12 at 23:28
    
It is Joseph Gallian: Contemporary Abstract Algebra, Seventh Edition, page 129 (chapter 6: Isomorphisms), Theorem 6.3. Picture to be attached to the original post within seconds. –  questionado Jan 18 '12 at 23:31

The inverse of an isomorphism is an isomorphism, so everything that works one way also works the other way as well. So either your reference is wrong, or you made an error in transcribing it. Did it really talk about iso-morphisms?

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Yes. In fact, my teacher made the same comment (though didn't have time to explain it in class). Maybe I/we are not seeing the "other direction" correctly, I am not sure what it would be. The context is abstract algebra/group theory. –  questionado Jan 18 '12 at 23:11

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