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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?

Watson
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  • I've seen this question few days ago (http://math.stackexchange.com/questions/278939/mathbbq-is-direct-product-of-two-non-trivial-subgroups). Well, maybe not posed in categorical terms. –  Jan 18 '13 at 09:03
  • 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? – SBF Jan 18 '13 at 09:08
  • 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 09:13
  • Also, I don't see how your proof wouldn't work for $\Bbb Q^2$ which instead can be represented as a product of two non-trivial groups. – SBF Jan 18 '13 at 09:15
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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. –  Jan 18 '13 at 09:16
  • @Ilya I'm not exactly sure, but I think the proof would break down for ${\mathbb Q}^{2}$ at the point where I use the fact that $\pi_{1}, \pi_{2}$ must be the zero map or inject an isomorphic copy into $G$ or $H$; I'm thinking this because if we do this the "hands on" way instead of the category theory way, we'd just be projecting one coordinate to $G$ and one coordinate to $H$. –  Jan 18 '13 at 09:18
  • Okay, now. I'm assuming that both $G$ and $H$ are not both trivial (this violates uniqueness of the middle mapping), so one of them has to be nontrivial. I picked $G\neq {0}$. From there, I'm assuming the left-hand triangle commutes, and I note that $\pi_{1}$ has to either be the zero map (which we can rule out easily) or inject an isomorphic copy of ${\mathbb Q}$ into $G$. At this point, I let $Z = G$ and let $f = id$. Then if $g\in G$ isn't part of the isomorphic copy of ${\mathbb Q}$ in G, then there cannot be a center map at all which makes the left-hand triangle commute. –  Jan 18 '13 at 09:23
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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
  • @Marek I think the only place where I am using a property of ${\mathbb Q}$ is in noting $\pi_{1},\pi_{2}$ need to either be the zero map or they need to inject an isomorphic copy of ${\mathbb Q}$ into $G$ or $H$. –  Jan 18 '13 at 19:46

3 Answers3

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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.

user26857
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Ittay Weiss
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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. –  Feb 27 '13 at 21:54
  • I'm glad you liked it :) – Ittay Weiss Feb 27 '13 at 21:56
  • Dear Ittay Weiss, in your solution above what do you mean by $\prod$ in "$\mathbb Q \cong G\prod H$"? Is this a typo or am I missing something here? – tcmtan Sep 25 '21 at 23:49
  • @tcmtan it's the categorical product of the two objects in the category $Ab$. – Ittay Weiss Sep 26 '21 at 10:54
  • Ah I see, thanks! I was not aware that there are different notations for the product in $\mathsf{Grp}$ and $\mathsf{Ab}$ because in Aluffi it also just uses $G\times H$ for product in $\mathsf{Ab}$ (at least insofar as I'm up to in the book, at which point this question asked by the OP was given). – tcmtan Sep 26 '21 at 11:24
  • It's not entirely standardised notation. Some authors use $\Pi$ for products in general, and $\times$ for finite products. – Ittay Weiss Sep 26 '21 at 15:36
  • Thank you for letting me know. – tcmtan Sep 26 '21 at 20:29
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Here's an attempt at a proof that is similar to Ittay Weiss':

In the category of abelian groups, saying that $\mathbb{Q}$ is a product is equivalent to saying that there exists $p : \mathbb{Q} \to \mathbb{Q}$, with $p^2 = p$. (For one direction: $\mathbb{Q} = \mathrm{ker} p \oplus \mathrm{im} p$)

$p(1) = \frac{n}{m}$ for some $n,m \in \mathbb{Z}, m \neq 0$.

$\frac{n}{m} = p(1) = p(p(1)) = p(\frac{n}{m}) = n p(\frac{1}{m})$

If $n=0$ then you can see that $p = 0$.

If $n \neq 0$ then $\frac{1}{m} = p(\frac{1}{m})$ so $p(1) = 1$

In this case $p$ must be $\mathrm{id}_\mathbb{Q}$.

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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.

Tunococ
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  • Thank you; this formalizes the first part. I don't think this immediately implies that $H$ is trivial, though --- –  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 00:59