There is a Hungerford problem to prove that the set of integers cannot be written as direct product of any family of its proper subgroups. In such questions, suppose if I move by the way of contradiction and take Z to be the direct product of its two proper subgroups, can I use that intersection of the proper subgroups should be trivial. My main doubt is: In such questions, one can think of two ways: internal direct product or the group is isomorphic to external direct product of its subgroups. Is it legitimate to use the first way as the question is not mentioning it ?

My second question is: are the two ways equivalent or in other words: if a group is isomorphic to external direct product of its subgroups, is it necessary that it is the internal weak direct product of some of its subgroups?

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    $\begingroup$ The internal product must be meant, because $\mathbb Z \cong 2\mathbb Z$, hence the integers are isomorphic to an external direct product of its proper subgroups (one factor only). $\endgroup$ – MooS Feb 27 '15 at 6:43
  • $\begingroup$ Of course the proof then comes down to the fact that two non-trivial subgroups of $\mathbb Z$ have non-trivial intersection. After that you could cover the external case with at least two factors and show that a direct product of infinite cyclic groups is never cyclic unless there is only factor. $\endgroup$ – MooS Feb 27 '15 at 7:24
  • $\begingroup$ What do you mean by "weak" direct product? If $\phi:G\rightarrow H_1\times H_2$ is an isomorphism of groups, then $G$ is the internal direct product of $\phi^{-1}(H_1\times 1)$ and $\phi^{-1}(1\times H_2)$. $\endgroup$ – Hagen Knaf Feb 27 '15 at 8:57

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