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The groups $(G, \cdot)$ and $(H, \odot)$ are isomorphic if a bijective map $\phi : G \rightarrow H$ exists, where $\phi (a \cdot b) = \phi(a) \odot \phi(b)$. I cannot find a map $\phi$ that does this when $G = \mathbb Z$ and $H = \mathbb Q$, and when $G = U(20)$ and $H = U(24) $. The group $U(n)$ uses the multiplication operation and comprises of all integers relatively prime to n. For example, $U(24) = \left\{ 1,5,7,11,13,17,19,23\right\}$ Is it correct to say that these groups are not isomorphic?

I thought of showing that U(20) and U(24) are of different order, but they have the same order. Thus, this proof failed.

MomoTheSir
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    $\Bbb Z$ and $\Bbb Q$ are not isomorphic. Nor are $U(20)$ and $U(24)$. But "I cannot find a map that does this" is not a proof... – anon Oct 24 '16 at 03:37
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    Note, your stated definition of isomorphic is incorrect. Not only should $\phi$ preserve the group operation, but it also needs to be bijective. In the definition you give $\phi$ is just a homomorphism. – wgrenard Oct 24 '16 at 03:39
  • Thanks for the feedback. I have added bijective into my definition of isomorphic. Also, how would I prove that two groups are not isomorphic? Thanks – MomoTheSir Oct 24 '16 at 03:43
  • This should help with the first one: http://math.stackexchange.com/questions/620551/prove-that-the-additive-groups-mathbbz-and-mathbbq-are-not-isomorphic – wgrenard Oct 24 '16 at 03:45

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The fact that you couldn't find a map is evidence that they might not be isomorphic, but that isn't a proof.

Here's an idea. You know that the integers with addition are a cyclic group. You know that an isomorphism maps a cyclic group to a cyclic group. If you can show the rationals under addition are not cyclic, you'll have your proof.

Nitin
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