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This is part of a question from Hungerford's Algebra text, specifically question 1 in section 1.8.

I came across a proof that says the following: suppose that $\mathbb{Z} = H_1 \times H_2$, where $H_1$ and $H_2$ are proper subgroups. Then $H_1 = \langle a \rangle$ and $H_2 = \langle b \rangle$ for $a, b \in \mathbb{Z}$. Then $H_1 \cap H_2 = \langle c \rangle$ for $c = lcm(a,b)$. This contradicts the fact that direct product components have trivial intersection.

But isn't this only true for internal direct products? Is it sufficient to show that $\mathbb{Z}$ is not an internal direct product of any proper subgroups for this problem? It kind of seems like it should be, since the proper subgroups are all normal, but I am somewhat confused. The components of a general direct sum are certainly allowed to have non-trivial intersections, e.g. $\mathbb{Z} \times \mathbb{Z}$.

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    $\begingroup$ Well, the thing is that the direct product $\Bbb Z\times\Bbb Z$ doesn't contain the two $\Bbb Z$s, but rather two isomorphic copies, $\{(n,0):n\in\Bbb Z\}$ and $\{(0,n):n\in\Bbb Z\}$, and these certainly have trivial intersection $\{(0,0)\}$. $\endgroup$ – Pedro Tamaroff Oct 26 '13 at 21:36
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    $\begingroup$ The components of a general direct sum are certainly not allowed to have nontrivial intersections. If $G=H\times K$ is a general direct sum, then $G$ has a copy of $H$ and $K$ sitting inside, and $G$ is an internal direct sum of $H$ and $K$, and conversely. Philosophically, the only difference between an internal direct sum and a general direct sum is whether you formed it as a sum to begin with or if you found out later it was a sum - so no structural difference at all. $\endgroup$ – anon Oct 26 '13 at 21:37
  • $\begingroup$ Okay, these are helpful comments, but for this proof that I found, if you consider $H_1$ to be $H_1 \times \{0\}$ and $H_2$ to be $\{0\} \times H_2$, don't they have a trivial intersection even if there is an element $(lcm(a,b)),0)$ in $H_1 \times \{0\}$ and there is an element $(0, lcm(a,b))$ in $\{0\} \times H_2$? $\endgroup$ – Athalianna Oct 26 '13 at 21:51
  • $\begingroup$ If $a\in H_1$ and $b\in H_2$ then the integer ${\rm lcm}(a,b)$ is a nonzero multiple of both $a$ and $b$ hence is in both $H_1$ and $H_2$. The internal definition of direct sum (trivial intersection, subgroup product is the whole group) does not reference coordinates. $\endgroup$ – anon Oct 26 '13 at 21:56
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    $\begingroup$ That is in the external definition. It is possible to form $H\times H$ for example, and $(a,0)$ and $(0,a)$ will be different elements. But we are talking about the internal definition. Here $H_1$ and $H_2$ are two subgroups of $\Bbb Z$. If we show $H_1$ and $H_2$ contain a nontrivial element of $\Bbb Z$, then $\Bbb Z$ cannot be an internal direct sum of them, because by definition $H_1\cap H_2$ must be trivial here. $\endgroup$ – anon Oct 26 '13 at 22:00
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I suspect you may have misunderstood the question. Although the subgroups $H_1,H_2 < G$ will intersect (in $0$ at least), we regard them as distinct when forming $H_1 \times H_2$. The problem here is to show that this cannot produce something isomorphic to $\mathbb{Z}$.

Hint: $\mathbb{Z}$ is cyclic.

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    $\begingroup$ There are cyclic groups which are direct products of some of its subgroups, so there has to be something else! $\endgroup$ – Mariano Suárez-Álvarez Oct 26 '13 at 21:50
  • $\begingroup$ Right, but in the case of $\mathbb{Z}$ its not too hard to see that the product of a family of its subgroups cannot be cyclic. $\endgroup$ – Elchanan Solomon Oct 26 '13 at 22:05
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    $\begingroup$ Well, that counts as something else :-) $\endgroup$ – Mariano Suárez-Álvarez Oct 26 '13 at 22:34

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