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Motivation + Problem: While doing some exercises in Aluffi's Algebra: Chapter 0, I came across a problem which asks the reader to prove that ${\mathbb Q}$ is not the product of two nontrivial groups. This is the standard product in the category of groups.

My Question: Since the chapter delves into categorical arguments, I tried it this way. We suppose not, that ${\mathbb Q} \cong G\times H$ and note that for any group $Z$ and any appropriate mappings $f,g$, we have the following diagram (sorry for the awkward TeXing, xypic doesn't seem to work here...),

$\displaystyle \begin{array}{ccccc} & & Z & & \\ & ^{f}\swarrow & \downarrow&_{\exists!\langle f,g\rangle} \searrow^{g} & \\ & G\longleftarrow_{\pi_{1}} & {\mathbb Q} & _{\pi_{2}}\longrightarrow H &\end{array}$

So, we find that there is always a unique mapping in the center if this is a product. The projections either inject an isomorphic copy of ${\mathbb Q}$ or are the zero mapping. We'd like to show that either $G$ or $H$ is trivial. I'm not sure how to show that at least one must be trivial.

My Attempt: Suppose $G \neq \{0\}$. We need to show $H = \{0\}$. Letting $Z = G$ and let $f = id$, we find that $G \cong {\mathbb Q}$. Moreover, the unique map in the center must be some multiplication mapping (which takes $x\mapsto qx$ for rational $q$; this is because $\pi_{1}$ must be a multiplication mapping if it is not the zero mapping). My guess here is that if $H$ is not trivial, it must also be an isomorphic copy of ${\mathbb Q}$ and we can allow $g$ to be some multiplication map like above such that the unique center mapping does not allow the right-hand triangle to commute.

Does this sort of argument work?

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I've seen this question few days ago (math.stackexchange.com/questions/278939/…). Well, maybe not posed in categorical terms. –  user26857 Jan 18 '13 at 9:03
    
I'm not an expert - just wonder how do you conclude that $G$ is isomorphic to $\Bbb Q$? Seems that you used the fact that $\pi_1\circ \langle \mathrm{id}_G,g\rangle = \mathrm{id}_G$, but it also works in case $G = \{0\}$, doesn't it? –  Ilya Jan 18 '13 at 9:08
    
I don't think you gain much from putting the problem in the language of category theory. The main idea of the proof is still the same. –  Tunococ Jan 18 '13 at 9:13
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Ah, I was typing up a response, but I'll type this first: @Tunococ, I'll agree that THIS proof can be done elegantly without a categorical argument, but I should be able to do this kind of thing categorically (eventually, with more complicated things!) and I'm stuck on even this simple case. –  james Jan 18 '13 at 9:16
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This seems like a very strange idea. To prove that $\mathbb Q$ is not a direct product, you must first and foremost use properties of $\mathbb Q$ itself. Can you point out where you are doing this? This kind of argument is necessary because there can't be any uniform categorical argument for general group since there are many groups which are direct products and many which aren't. On the other hand, I don't see any advantage in translating the properties of $\mathbb Q$ (e.g. characterizing it by some universal property) to categorical language. –  Marek Jan 18 '13 at 11:18
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2 Answers

up vote 3 down vote accepted

If $\mathbb Q$ was isomorphic to $G\times H$ then $G$ and $H$ must be abelian and so $\mathbb Q$ would be a product $\mathbb Q \cong G\prod H$in the category $Ab$, and as in $Ab$ finite products and coproducts agree, $\mathbb Q \cong G\coprod H$ holds as well. Let $i:G\to \mathbb Q$ and $j:H\to \mathbb Q$ be the canonical injections. In $Ab$ the canonical injections are monos and monos are injective functions. So, $G$ and $H$ are canonically subobjects of $\mathbb Q$. Now, the special property of $\mathbb Q$ is that the intersection of any two non-trivial subobjects in it have a non-trivial intersection (with $\mathbb Z$). However, in $Ab$ the canonical injections $G\to G\coprod H$ and $H\to G\coprod H$ intersect at the $0$ object, contradiction.

This is about as categorically as I could furnish the proof, I hope you like it.

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I enjoyed this one quite a bit! The only thing I've got to parse out for myself is the "special property" of ${\mathbb Q}$, but this is exactly what I wanted. –  james Feb 27 '13 at 21:54
    
I'm glad you liked it :) –  Ittay Weiss Feb 27 '13 at 21:56
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This is my attempt at guessing what you're looking for. I still don't think it is a good thing to do. No new ideas can come out of this. The proof just looks unnecessarily hard to read. (There are facts used here are specific to the category of groups that I simply assume without mentioning.)

According to you diagram, suppose $\iota: G \to \mathbb Q$ is injective, i.e., $G$ is a subobject of $\mathbb Q$.

Define $\tilde\pi_1 = \iota \circ \pi_1: \mathbb Q \to \mathbb Q$. For any integers $a, b$ with $b \ne 0$, $$\tilde\pi_1(\frac ab) = \frac 1bb \tilde\pi_1(\frac ab) = \frac 1b \tilde\pi_1(b \frac ab) = \frac 1b \tilde\pi_1(a) = \frac ab \tilde\pi_1(1).$$ This means $\tilde\pi_1$ is simply multiplication by $\tilde\pi_1(1)$.

If $\tilde\pi_1(1) = 0$, then $\tilde\pi_1 = 0$, and since $\iota$ is injective, we must have $\pi_1 = 0$. This forces $G$ to be the zero object (because for any choice of $Z$, $f:Z \to G$ must be a zero morphism because it factors through $\pi_1$).

Otherwise $\tilde\pi_1(1) \ne 0$. From the equations above, $\tilde\pi_1$ is surjective. (Given $q \in \mathbb Q$, pick $p = q / \tilde\pi_1(1)$. Then $\tilde\pi_1(p) = q$.) It is also injective because $\tilde\pi_1(q) = 0$ implies $q = 0$. So $\tilde\pi_1$ is an isomorphism $\Rightarrow$ $\iota$ is surjective and $\pi_1$ is injective $\Rightarrow$ $\iota$ is an isomorphism $\Rightarrow$ $\pi_1$ is an isomorphism.

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Thank you; this formalizes the first part. I don't think this immediately implies that $H$ is trivial, though --- –  james Jan 18 '13 at 19:54
    
In the category of groups, $G$ and $H$ have zero intersection in $G \times H$. –  Tunococ Jan 19 '13 at 0:59
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