5

I am trying to check if the following relationship is correct. Let $F$ be a free group over some set $X$. Let $\alpha_1, \alpha_2, ..., \alpha_n$ be epimorphisms from $F$ to groups $H_1, H_2, ..., H_n$ respectively. Let $K_1, K_2, ..., K_n$ be the kernels of $\alpha_1, \alpha_2, ..., \alpha_n$ respectively. It follows that $K_1, K_2, ..., K_n$ are all normal.

Now, let $K= \bigcap K_i$ for $i=1,...,n$ (i.e. the normal group formed from the intersection of the kernels). We can now form the quotient $F/K$.

Is it true that $F/K$ is isomorphic to a subcartesian product of $F/K_1,F/K_2,...,F/K_n$ ? How could I show it ?

I want to check if this is correct since this would then mean that $F/K$ is also isomorphic to a subcartesian product of $H_1,H_2,...,H_n$ ...

(For context over this question, am trying to understand a proof for Birkhoff's variety theorem which uses the above)

Shaun
  • 44,997
Link L
  • 717

2 Answers2

3

Recall that a subdirect product of $\{G_i\}_{i\in I}$ is a subgroup $M\leq \prod_{i\in I}G_i$ such that $\pi_j(M)=G_j$ for all $j\in I$.

In general, if $G$ is a group, $\{H_i\}_{i\in I}$ is a family of groups, and $f_i\colon G\to H_i$ are morphisms, then we obtain an induced homomorphism $$f\colon G\to \prod_{i\in I}H_i$$ by the universal property of the product. This map has the property that for each $i\in I$, $\pi_i\circ f = f_i$. If $I=\{1,\ldots,n\}$, you can "visualize" $f$ as the map $f(g) = (f_1(g),\ldots,f_n(g))$.

In this case, the image of $f$ is always a subdirect product of the family $\{f_i(G)\}_{i\in I}$.

Indeed, given $h_i\in f_i(G)$, we know there exists $g\in G$ such that $f_i(g)=h_i$. Then $\pi_i(f(g)) = f_i(g) = h_i$ by definition, so there exists an element in $\mathrm{Im}(f)$ that has $i$th coordinate equal to $h_i$. This proves that $f(G)$ is a subdirect product of $\{f_i(G)\}_{i\in I} = \{\mathrm{Im}(f_i)\}_{i\in I}$.

In your situation, $G=F$ is a free group and the maps $\alpha_i$ are surjective (I avoid "epimorphism" because there are varieties of groups where you have nonsurjective epimorphisms; of course, in the category of all groups epimorphisms are surjective, but I prefer not to conflate the two). So certainly the map $F\to H_1\times\cdots\times H_n$ has image that is a subdirect product. The only thing remaining is to verify that the morphism factors through $K$, but that follows because the kernel of the universal map $\alpha\colon F\to H_1\times\cdots\times H_n$ is precisely the intersection of the kernels of the maps $\alpha_i$, which is $K$.

Arturo Magidin
  • 398,050
2

Edit: as pointed out in the comments, "subcartesian product" has a formal meaning and is not the same as a cartesian product of subgroups.

Some thoughts which I found too long for a comment:

The maps $\alpha_i$ define a map

$$ f \colon t \in F \mapsto (\alpha_1(t),\ldots, \alpha_n(t)) \in H_1 \times \cdots \times H_n $$

Its kernel consists of the $t \in F$ for which $\alpha_i(t) = 1$ for all $i$ simultaneously, i.e. $K = \cap_i K_i$. Hence

$$ F/K \simeq \mathbf{im} f \leq H_1 \times \cdots \times H_n. $$

We have not used that each $\alpha_i$ is epi nor the fact that $F$ is free. In general, the image of $f$ need not be a cartesian product of subgroups, even with these hypotheses (e.g. if $F = \Bbb Z$, $n = 2$, and $\alpha_1 = \alpha_2 = id$, then $f$ is the diagonal map $\Delta \colon \Bbb Z \to \Bbb Z^2$).

It may be that $F/K$ is still isomorphic to some cartesian product through some other morphism.

qualcuno
  • 17,121
  • I am not sure to understand your example. One has $\Delta(f) = {(n,n) \mid n \in {\Bbb Z}}$. Isn't it a subgroup of ${\Bbb Z}^2$ which maps onto ${\Bbb Z}$ under the two projections? – J.-E. Pin May 20 '21 at 18:15
  • Do you mean $\Delta(Id)$? Yes, indeed. I didn't understand your confusion, care to elaborate? – qualcuno May 20 '21 at 18:49
  • Sorry, I mean $\Delta({\Bbb Z})$. – J.-E. Pin May 20 '21 at 19:34
  • Ok, but I don't understand what the confusion is/where it stems from – qualcuno May 20 '21 at 20:21
  • Why do you say that the image of $f$ is not a subcartesian product? – J.-E. Pin May 20 '21 at 21:12
  • In that case, the diagonal is isomorphic to $\mathbb Z$, my point is that that need not be the case in general. A subgroup of $H \times G$ need not be of the the form $H' \times G'$ with $H \leq H', G \leq G'$. The diagonal in $\mathbb Z^2$ is such an example: there are no $m\mathbb Z, n \mathbb Z$ such that $\Delta(\mathbb Z) = m \mathbb Z \times n\mathbb Z$. As the posts says, it may still be that the image of $f$ is isomorphic to some cartesian product. – qualcuno May 20 '21 at 21:19
  • 1
    I am afraid that your definition of a subcartesian product is not right. – J.-E. Pin May 20 '21 at 21:23
  • @J.-E.Pin I was unaware that this had a formal meaning. Fixed to convey what I originally meant! Thanks. I can also see that the question meant something different... Also, what's the intuition for this? I would expect a subcartesian product to be, in general, a non-trivial product of groups in of itself. But the definition suggests otherwise, I think – qualcuno May 20 '21 at 21:31
  • 1
    @guidoar: Subdirect products are an important tool when dealing with decompositions of algebras in general. In the case of products of two groups, you may be familiar with Goursat's Lemma which uses subdirect products to describe all subgroups of a product $G\times H$. – Arturo Magidin May 20 '21 at 21:44
  • I see! Thanks for the link Arturo, I will give it a read; I did not know about Goursat's Lemma (I think the only result in that direction that I can recall is that if $G \times H$ is finite with factors of coprime order then its subgroups are actually product of subgroups). – qualcuno May 20 '21 at 21:55
  • 1
    @guidoar: That would follow from Goursat's Lemma, since in that case the isomorphisms that can occur are trivial. – Arturo Magidin May 20 '21 at 22:09
  • 1
    @guidoar: See, e.g., this answer. – Arturo Magidin May 20 '21 at 22:12