# How can I prove that the additive group of rationals is not isomorphic to a direct product of two nontrivial groups?

I am working through Paolo Aluffi's new GSM text on my own (self-study). On page 63, he asks the reader to

Prove that $\Bbb{Q}$ is not the direct product of two nontrivial groups.

For some context, this is an exercise following a section entitled "The category Grp". I am assuming that he means "is not isomorphic to the direct product of two nontrivial groups", and I can see two possible ways to proceed with this proof, but have been unsuccessful with either approach.

Approach 1: Show that the additive group of rationals has a property that is preserved by isomorphism that the direct product of two nontrivial groups does not have or vice-versa. This seems challenging unless I can significantly narrow down the properties that a direct product of two nontrivial groups that was isomorphic to $\Bbb{Q}$ would necessarily have.

Approach 2: Considering the section in which this question occurs, show that if $G$ and $H$ are nontrivial groups and $\Bbb{Q} \cong G \times H$, then there are homomorphisms $\varphi_{G}:\Bbb{Q} \rightarrow G$ and $\varphi_{H}:\Bbb{Q} \rightarrow H$ which do not factor or do not factor uniquely through the product $G \times H$. This would be a contradiction, as $G \times H$ is a final object in the category Grp.

I would greatly appreciate suggestions on how to proceed further with either of these approaches or with alternate approaches.

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For any $p,q\in \mathbb Q\setminus 0$, there are non-zero integers $n$ and $m$ such that $np = mq\neq 0$. That is not true for the product of two non-trivial groups. – Thomas Andrews Dec 19 '11 at 22:02

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Is there more than one David Pincus, or do I have two of your papers sitting on my desk? – Asaf Karagila Dec 19 '11 at 22:05
IOW Thomas' hint goes together with your Approach #1: A direct product has a pair of non-trivial subgroups that intersects trivially whereas – Jyrki Lahtonen Dec 19 '11 at 22:05
Here is an approach: show that any two nontrivial subgroups of $\mathbb{Q}$ have nontrivial intersection. Seems very localized though, I would like to see a more general argument. – François G. Dorais Dec 19 '11 at 22:07

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Isomorphisms are, by definition, homomorphism that have an inverse that is also a homomorphism. So, yes, isomorphism is, by definition, assumed to be a homomorphism. – Arturo Magidin Mar 30 '12 at 18:20
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## 1 Answer

The endomorphism ring of a direct product is never a domain, yet the endomorphism ring of ℚ is a field.

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What is the endomorphism ring isomorphic to? To itself? I thought about the fact that $\mathbb Q$ cannot be isomorphic to a direct product of two rings (because one has non-trivial ideals and $\mathbb Q$ does not), but didn't know how to use that property. – Patrick Da Silva Dec 19 '11 at 22:11
@PatrickDaSilva, indeed, the endomorphism ring of the abelian group $\mathbb Q$ is isomorphic to ring $\mathbb Q$. – Mariano Suárez-Alvarez♦ Dec 19 '11 at 22:13
So my idea wasn't so bad after all. Thanks =) Perhaps I am not used to look at the endomorphism ring of a group to study its structure. =P – Patrick Da Silva Dec 19 '11 at 22:14

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@Patrick: The simple way to establish this is to show that if you know what $1$ maps to, then you know where everything maps to, and that $1$ can map anywhere. $\mathbb{Q}$ is a "free torsionfree divisible group of rank 1". – Arturo Magidin Mar 30 '12 at 18:22
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