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Stream: learning: questions

Topic: Algebra (Aluffi) exercises

Jonathan Castello (Mar 27 2020 at 07:19):

I'm working through Aluffi's Algebra: Chapter Zero to pick up category theory on my second pass through abstract algebra, so I think this would be considered a "basic question"...

Chapter II, Exercise 3.5 asks the reader to prove that $\mathbb{Q}$ is not the direct product of two non-trivial groups. (Because it's obviously isomorphic to $\mathbb{Q} \times \mathbb{1}$.) I've found a couple scattered proofs online that I can follow, but it isn't clear at all why I should expect this to be true. It feels like the proof is a bag of tricks that happens to work, and that's not very satisfying.

Complicating matters is that Aluffi hasn't even introduced basic concepts like "subgroup" or "kernel" at this point, so I don't feel licensed to apply them directly. But this exercise is in the first chapter introducing the category Grp, so I feel obliged to find a more categorial solution.

Here's the proof I have written down right now. It's slightly adapted from this MathOverflow question. Is there a "better" way to see that this theorem holds?

1. Suppose, for the sake of contradiction, that $\mathbb{Q} \cong G \times H$ for some non-trivial groups $G, H$, with projections $\pi_G, \pi_H$.
2. Observe that 0 is the unique element of $\mathbb{Q}$ with finite order.
3. Observe that, for every rational $\frac{m}{n} \in \mathbb{Q}$, we have $(\pi_G(\frac{m}{n}))^n = \pi_G(m) = (\pi_G(1))^m$.
4. Supposing that $\pi_G(\frac{m}{n}) = e_G$ for some non-zero rational, we determine that $\pi_G(\frac{m}{n}) = e_G$ for every rational by application of the above observations.
5. Therefore, $\pi_G$ is the constant function to zero, forcing $G$ to be trivial. By hypothesis, it is not, so the fiber of 0 under $\pi_G$ is a singleton.
6. But since $\pi_G((e_G, h)) = e_G$ for any $h$, there must be only a single possible $h$, and hence $H$ is trivial, again a contradiction.

Jonathan Castello (Mar 27 2020 at 07:33):

(What's especially frustrating is that the previous exercise, Ch II E 3.4, is an absolute gem of a puzzle -- finding non-trivial groups $G, H$ such that $G \cong H \times G$. To go from that to this is... :sob:)

Reid Barton (Mar 27 2020 at 15:32):

Any group $G \times H$ with $G$ and $H$ nontrivial has two nontrivial subgroups $G \times 1$, $1 \times H$ which intersect only in the identity. But $\mathbb{Q}$ doesn't have this property (because any two non-identity elements have a common multiple/power).

John Baez (Mar 27 2020 at 19:35):

@Jonathan Castello - the proof you give seems simple enough, but we can clean it up a little and make it shine. It's enough to prove $G$ is trivial - you're trying to show that if $\mathbb{Q}$ is a product of groups one of them must be trivial - so you can stop after step 4.

Jonathan Castello (Mar 27 2020 at 19:58):

John Baez said:

Jonathan Castello - the proof you give seems simple enough, but we can clean it up a little and make it shine. It's enough to prove $G$ is trivial - you're trying to show that if $\mathbb{Q}$ is a product of groups one of them must be trivial - so you can stop after step 4.

Step 4 introduced a hypothetical context, though: assume that $\pi_G$ maps some non-zero rational to the identity. So at that point, I've only invalidated that hypothesis, not the product decomposition hypothesis. No?

Jonathan Castello (Mar 27 2020 at 20:03):

Reid Barton said:

Any group $G \times H$ with $G$ and $H$ nontrivial has two nontrivial subgroups $G \times 1$, $1 \times H$ which intersect only in the identity. But $\mathbb{Q}$ doesn't have this property (because any two non-identity elements have a common multiple/power).

Right, I think I follow that argument. Unfortunately, subgroups haven't been introduced at this point in the book, so I feel a little dirty invoking them for this exercise. Feels like this proof is dancing around that property though -- maybe I can unwrap it a bit...

Jonathan Castello (Mar 27 2020 at 20:11):

Jonathan Castello said:

Step 4 introduced a hypothetical context, though: assume that $\pi_G$ maps some non-zero rational to the identity.

Ahhhh, wait -- I can apply the logic from step 6 to obtain this as a fact instead of as a hypothesis. Every $q \cong (e_G, h)$ maps to $e_G$, and only one of them can be 0.

Jonathan Castello (Mar 27 2020 at 20:18):

I'm much happier with this proof now! I'm leaving out some mechanical parts to focus on the structure, and it feels a lot better now. It leads with a small series of lemmas about $Q$ (building up to observation 4), then applies them to immediately get the main result.

1. Let $G, H$ be groups such that $\mathbb{Q} \cong G \times H$, with projections $\pi_G, \pi_H$.
2. Observe that 0 is the unique element of $\mathbb{Q}$ with finite order.
3. Observe that, for every rational $\frac{m}{n} \in \mathbb{Q}$, we have $(\pi_G(\frac{m}{n}))^n = (\pi_G(1))^m$.
4. Observe that, if $\pi_G(\frac{m}{n}) = e_G$ for some non-zero rational, then $\pi_G(\frac{m}{n}) = e_G$ for every rational by application of the above observations.
5. Since $\pi_G((e_G, h)) = e_G$ for any $h \in H$, and since only one $(e_G, h)$ can be $0 \in \mathbb{Q}$, by observation 4 we see that either $G$ or $H$ must be trivial.

John Baez (Mar 27 2020 at 20:18):

Let's see how I'd prove the result in Aluffi. I think this is more pleasant if we introduce some terminology and concepts. Let me use additive notation for group operations since all the groups involved must be abelian: since $\mathbb{Q}$ is, so must be $G$ and $H$.

Given a group element $a$ let's call $a$ added to itself $n$ times for $n = 1, 2, 3...$ a multiple of $a$.

A group is torsion-free if when any multiple of an element is zero, that element must be zero.

$\mathbb{Q}$ is torsion-free. Any subgroup of a torsion-free group is torsion-free.

Now assume $\mathbb{Q} \cong G \times H$. We wish to show either $G$ or $H$ is trivial. To do this let's assume $H$ is nontrivial and show $G$ is trivial.

Since $G$ and $H$ are isomorphic to subgroups of $\mathbb{Q}$ they must be torsion-free.

Let $\pi_G \colon \mathbb{Q} \to G$ be the projection onto $G$.

Since $H$ is nontrivial there must be some nonzero element $q \in \mathbb{Q}$ with $\pi_G(q) = 0$. This implies that $\pi_G$ of any multiple of $q$ is also zero. But for every $q' \in \mathbb{Q}$ we can find some multiple of $q'$ that is also a multiple of $q$. So, for every $q' \in \mathbb{Q}$, $\pi_G$ of some multiple of $q'$ is zero. Since $G$ is torsion-free this implies that $\pi_G(q') = 0$.

In other words, $\pi_G$ maps everything to zero. So $G$ must be trivial!

Jonathan Castello (Mar 27 2020 at 20:27):

Very nice! The use of "subgroup" here is more obviously benign to me, since the torsion-free property clearly applies to all elements regardless of what bin we're putting them in -- it's just conventional nomenclature. (We're not using additive closure, specifically.)

John Baez (Mar 27 2020 at 20:52):

Okay, good. To me reasoning about products of groups without mentioning subgroups is like boxing with one hand tied behind one's back, since one great thing about a product of groups $G \times H$ is that it has homomorphisms to $G$ and $H$ but there are also homomorphisms from $G$ and $H$ back into it, which reveal them as subgroups of $G \times H$.

But, if you're working through a textbook, I guess it's a nice exercise to solve the problems using the concepts that you're supposed to be allowed to use at a given moment. Sometimes boxing with one hand behind one's back can be good exercise....

... but you can get beaten up.

Reid Barton (Mar 27 2020 at 20:55):

I don't know what the author had in mind, but I would suggest the lesson to be learned here is sometimes it's better not to tie one hand behind your back.

Jonathan Castello (Mar 27 2020 at 20:55):

Haha, yes, exactly! If the book just introduced some concepts, I'd like to apply them specifically :wink:

The point about having both projections and injections does come up in the chapter, as in $Ab$ products align with coproducts. I think that, especially, calling out "subgroups" explicitly feel unnecessary in this context -- the injections and projections are both pretty obvious here.

(I think you still get injections into $G \times H$ in $Grp$, you just don't get the universal property of coproducts? But all groups involved in this problem are abelian anyway.)

Jonathan Castello (Mar 27 2020 at 21:01):

@Reid Barton, if I'm being honest, I'm too out of practice with algebra to be able to confidently recall properties like the subgroup intersection one you proposed. My strategy of using only what the book has provided so far is as much a pedagogical tool as a way of not leading myself down paths I can't navigate yet.

(Just about the only interesting thing I can really confidently remember is that you can quotient by normal subgroups, and what a normal subgroup is. Undergrad abstract algebra was more than five years ago, and I haven't had a lot of opportunities to practice my group theory since then.)

John Baez (Mar 27 2020 at 21:05):

Yes, I was just thinking about that. Here is a little puzzle for the category theorists out there:

Puzzle. In the category $\mathsf{Gp}$ binary products have a special feature: besides the usual projections $\pi_1 \colon G \times H \to G$ and $\pi_2 \colon G \times H \to G$, we also have morphisms $i_1 \colon G \to G \times H$ and $i_2 \colon H \to G \times H$. In $\mathsf{AbGp}$ this arises because binary products are also binary coproducts, but in $\mathsf{Gp}$ they are not. So, why do we have these morphisms $i_1 \colon G \to G \times H$ and $i_2 \colon H \to G \times H$ in $\mathsf{Gp}$?

There are many levels of answering this question "why". You can just write down the formula for these morphisms, but that's boring. A better style of answering the question would be to say something about in which kind of categories this phenomenon happens.

For starters, what are some other categories where it happens, even though binary products aren't binary coproducts?

John Baez (Mar 27 2020 at 21:08):

@Joe Moeller loves this puzzle, but for that reason he's not allowed to answer it. :smiling_devil:

Mike Stay (Mar 27 2020 at 21:20):

Does it have to do with the fact that groups are pointed, so we can take any g in G to (g, 1) in G x H?

Jonathan Castello (Mar 27 2020 at 21:22):

:thinking: Trivial groups are both initial and final in $\mathsf{Gp}$ (the book calls them "zero objects"). So for any group $G$, we have unique morphisms $a_G : \mathbb{1} \to G$ and $z_G : G \to \mathbb{1}$.

Now, for any product $G \times H$ we may construct the left injection $\langle id_G, a_H z_G \rangle : G \to G \times H$, and similarly for the right injection.

So I would wager that products have injections in any category with zero objects.

T Murrills (Mar 27 2020 at 21:22):

Hmmm...as a first stab at it, we seem to recover these maps as the product of the identity map on the relevant component with the unique map out of the category’s initial object and into the other component!

However, this actually gives a map from the product of the relevant component with the initial object. So, our next task is characterizing the conditions under which any G is canonically isomorphic to its product with the initial object... (is that true in general, maybe? hmm...)

David Egolf (Mar 27 2020 at 21:24):

Digging out my copy of Aluffi, it appears that binary products and binary coproducts coincide also in the category of $R$-modules. However, I don't (yet) have any intuition as to why this is so. So I guess we're interested in categories where we don't have this coincidence, but we still have morphisms into the product.

Jonathan Castello (Mar 27 2020 at 21:28):

My $a_G$ is exactly the pointed element @Mike Stay mentioned, so perhaps it would have been better to call it $e_G$ -- but I was going with an alphabetical mnemonic for "initial" and "final" :laughing:

Stelios Tsampas (Mar 27 2020 at 21:29):

John Baez said:

There are many levels of answering this question "why". You can just write down the formula for these morphisms, but that's boring. A better style of answering the question would be to say something about in which kind of categories this phenomenon happens.

OK, here's my limp: The projections are actually counits for the "writer" comonad and the injections are units for the "writer" monad, as the groups involved have the "monoidal" requirements for it. In the self-dual category $Rel$ of relations products and coproducts coincide, so i have a feeling sefl-duality has something to do with what's been happening here.

vikraman (Mar 27 2020 at 21:29):

With a zero object we always get $X \rightarrow 0 \rightarrow Y$.

Stelios Tsampas (Mar 27 2020 at 21:29):

Of course, @Mike Stay has already pointed out that groups are pointed, a.k.a. the "monoidal" requirement I mention.

T Murrills (Mar 27 2020 at 21:32):

ah, okay, I think @Jonathan Castello has it: we simply need a diagonal map from G to GxG, which we’re guaranteed by, well, the universal property of products.

But are there cases where the projection from the product of anything with the initial object to the nontrivial object is an isomorphism, and the initial object is not a zero object? Presumably the injections would still exist in that more general setting.

Jonathan Castello (Mar 27 2020 at 21:51):

:thinking: @T Murrills, consider a poset with greatest element $\top$ and binary meets $\wedge$ (greatest lower bounds). This poset is a monoidal category, but its products do not have injections. (It has no loops at all, other than identity arrows.)

(In particular, $x \wedge y \le x$ and $x \wedge y \le y$ are the "projections", but we cannot expect $x \le x \wedge y$.)

T Murrills (Mar 27 2020 at 22:07):

@Jonathan Castello That example seems a little funky somehow. If arrows go from smaller to greater, as indicated by the direction of the projections, it seems the greatest element is a terminal object, not an initial one. So, a smallest element would be an initial object, and the product of x with the initial object is the initial object, and thus not isomorphic to x (and so doesn’t fall in the scope of what I’m considering). Is this right, or have I flipped something?

Jonathan Castello (Mar 27 2020 at 22:07):

No, you're right... :thinking:

David Egolf (Mar 27 2020 at 22:15):

Let $\mathbf{C}$ be a category with a binary product. So, for any objects $G,H$ in $\mathbf{C}$ we can form a product $G \times H$. Then, by the universal property for products, there is a morphism from $G$ to $G \times H$ if there exists a morphism $f_1: G \to G$ and a morphism $f_2: G \to H$. This morphism (I think) is the "injection morphism" $i_G:G \to G \times H$ in the case that $f_1 = id_G$. Note that there always exists at least one morphism from $G$ to $G$ and also at least one from $H$ to $H$, thanks to the identity morphisms. So, it seems that we have these injection morphisms for both $G$ and $H$ when there is a morphism from $G$ to $H$ and a morphism from $H$ to $G$. Note that if we have a zero object $0$ then we can always make a morphism from $G$ to $H$ by composing the unique morphism from $G$ to $0$ with the unique morphism from $0$ to $H$. Similarly, we can always make a morphism from $H$ to $G$ if there is a zero object.

So, I think these injection morphisms exist for any two objects $G$ and $H$ as long as in the category there is at least one morphism from $G$ to $H$ and at least one morphism from $H$ to $G$, no matter what the two objects are.

(I don't know much category theory... am hoping the above is not horribly wrong. Any corrections appreciated!)

Jonathan Castello (Mar 27 2020 at 22:19):

@David Egolf, is it the case that $\pi_G i_G = {id}_G$ in your construction? I think you've described the pairing property we get directly from products, but injection comes with additional expectations about the arrow we get.

David Egolf (Mar 27 2020 at 22:23):

@Jonathan Castello Yes, $\pi_ G i_G = id_G$ in this construction.
Ah, I think I see what you mean. You're pointing out that we don't just care about the existence of morphisms $i_1: G \to G \times H$ and $i_2: H \to G \times H$ but we want these morphisms to be nice in some sense.
I guess the stuff I said above is necessary for these injection morphisms to exist, but (probably?) not sufficient.

Jonathan Castello (Mar 27 2020 at 22:26):

I don't see how this construction forces that equality. Certainly the arrow you're pairing on the left is the identity, but the arrow on the right is arbitrary. Unless the pairing operator is constant over the right operand (i.e. it just ignores it), it seems that the paired arrow ought to assign different elements of $G \times H$ to each $G$ depending on what your choice of right function is.

(And the pairing operator can't be constant, or it wouldn't support the projection onto of the right component of the product.)

David Egolf (Mar 27 2020 at 22:34):

I'm not quite following you.... I'm afraid I'm not familiar with the terminology of "pairing". I'd be happy to learn about it though!

In some more detail, here is why I think this construction forces that equality:
Let us fix a product $G \times H$ with projections $\pi_G$ and $\pi_H$. Then I select $G$ and choose the identity morphism from $G$ to itself and choose the morphism $f$ from $G$ to $H$. Then by the universal property of products there is a unique morphism $\sigma: G \to G \times H$ so that $\pi_G \sigma = id_G$ and $\pi_H \sigma = f$. I called $\sigma$ by the name $i_G$ above.

David Egolf (Mar 27 2020 at 22:36):

Ah, the terminology $i_G$ is bad and the mapping $\sigma$ is not the injection. It has to also include information about $f$. (I think this is what you mean by pairing?). However, we do still have a morphism from $G$ to $G \times H$.

Jonathan Castello (Mar 27 2020 at 22:36):

"Pairing" is obtaining $\sigma$ such that composition with the left projection gives you the left function you put in, and composition with the right projection gives you the right function you put in. We can write $\sigma = \langle id_G, f \rangle$ to make that property explicit (which should also graphically motivate the term "pairing").

Jonathan Castello (Mar 27 2020 at 22:38):

Yes, exactly -- your injection depends on the other function :grinning: but we want a single injection that works in general.

John Baez (Mar 27 2020 at 22:38):

Mike wrote:

Does it have to do with the fact that groups are pointed, so we can take any g in G to (g, 1) in G x H?

Yes, it has a lot to do with that fact! Can you say super-precisely what you mean by "pointed" here? That would take me close to the best answer I know to this puzzle. Of course the identity element of a group is a point in the usual set-theoretic sense, but we want something more category-theoretic, so we can take any category and guess when products will have these 'inclusions' as well as the usual projections.

John Baez (Mar 27 2020 at 22:40):

I see now that lots of other people have answered this puzzle, going further in the category-theoretic direction.

Jonathan Castello (Mar 27 2020 at 22:40):

@T Murrills Here's my attempt at resolving your question: If you have a categorical product with an identity, then the identity must be final.

First, there exists some arrow $G \to \mathbb{1}$, since $G \times \mathbb{1}$ has a projection to $\mathbb{1}$ and $G \cong G \times \mathbb{1}$.

Second, this arrow must be unique: take any two arrows $G \to \mathbb{1}$ and pair them to get an arrow $G \to \mathbb{1} \times \mathbb{1}$. Well, $\mathbb{1} \times \mathbb{1} \cong \mathbb{1}$, and the projections out of $\mathbb{1}$ force these arrows to be the same. (A bit hand-wavey, here :thinking:)

This is at least intuitively motivated by the idea that the identity is the empty product. So it ought to satisfy the same flavor of universal property that the product itself does.

John Baez (Mar 27 2020 at 22:46):

@Jonathan Castello wrote:

So I would wager that products have injections in any category with zero objects.

Yes, that's the answer I was thinking about! Just to remind everyone, a zero object is an object that's initial and terminal. Whenever we have a zero object, there's always a 'zero morphism' from any object $X$ to any object $Y$, namely the composite $X \to 0 \to Y$ where $0$ is the zero object. We call this morphism $0 \colon X \to Y$.

Thus, in a category with a zero object, a product $A \times B$ comes with morphisms $i_1 \colon A \to A \times B$, $i_2 \colon B \to A \times B$. To get $i_1$ we just take the pairing of $1 \colon A \to A$ and $0 \colon A \to B$. Similarly for $i_2$.

John Baez (Mar 27 2020 at 22:49):

Puzzle. In this situation, is it always true that $i_1 \colon A \to A \times B$ composed with $\pi_1 \colon A \times B \to A$ is the identity?

Jonathan Castello (Mar 27 2020 at 22:54):

Isn't that immediate from the universal property of products? That $\pi_1 \langle f, g \rangle = f$? We defined $i_1 = \langle 1, 0 \rangle$.

John Baez (Mar 27 2020 at 22:55):

Yes, it's immediate.

David Egolf (Mar 27 2020 at 22:56):

Ok, that makes sense. To make my $i_G$ behave actually like a (unique) injection, we need to set $f$ uniquely. To do that, we can use the special zero morphism that exists between each pair of objects when we have a zero object. Further, we use the zero morphism to make our injection toss out all information about $H$, which is an important part of what we want an injection to do.

John Baez (Mar 27 2020 at 22:58):

Only for those who don't find it "immediate":

Puzzle. Show that in a category with a zero object, the composite of any morphism with a zero morphism is a zero morphism.

John Baez (Mar 27 2020 at 22:59):

So, the zero morphisms form a kind of "ideal" in the category.

Jonathan Castello (Mar 27 2020 at 23:01):

Hahahaha, I was just thinking that about that :sweat_smile: In $\mathsf{Gp}$ we have the obvious trivial homomorphisms, and since we know that the kernel of a group homomorphism is a normal subgroup of the image, we literally have an ideal.

Jonathan Castello (Mar 27 2020 at 23:02):

Oh wait, I see, this is an even more interesting kind of ideal!

John Baez (Mar 27 2020 at 23:09):

I just meant an ideal in this sense: if f is a morphism in the ideal and g, h are any morphisms that can be composed with f on the left and right, gf and fh are in the ideal.

Jonathan Castello (Mar 27 2020 at 23:10):

Yes, I got a little excited thinking about groups specifically, and failed to realize that this is actually a matter of the morphisms, not of the objects. :sweat_smile:

John Baez (Mar 27 2020 at 23:10):

For example any ring gives a one-object category where the morphisms are ring elements and composition is multiplication; then the kind of ideal I'm talking about reduces to the usual kind of ideal in ring theory.

Nathanael Arkor (Mar 27 2020 at 23:13):

is this some sort of "bisieve"?

John Baez (Mar 27 2020 at 23:18):

I guess so! I guess a "sieve" is a "right ideal".

David Egolf (Mar 27 2020 at 23:33):

Ok, here's a stab at showing that in a category with a zero object $0$ composing with a zero morphism yields a zero morphism.

Let $G, K, H$ be objects. Let $!_{G \to 0}$ be the unique morphism from $G$ to $0$ and $!_{0 \to K}$ be the unique morphism from $0$ to $K$. Finally, let $f$ be some morphism $f: K \to H$.

Create a zero morphism going from $G$ to $K$ by $!_{0 \to K} \circ !_{G \to 0}$. Note that $f \circ !_{0 \to K} = !_{0 \to H}$ because there is only one morphism from $0$ to $H$. We now compose $f$ with the zero morphism formed earlier, getting $f \circ !_{0 \to K} \circ !_{G \to 0} = !_{0 \to H} \circ !_{G \to 0}$, which is a zero morphism.

I think the argument will be similar when composing with a zero morphism on the other side.

John Baez (Mar 27 2020 at 23:38):

Yes, that's right! :+1: