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Stream: learning: questions

Topic: factorization systems of classical groups

John Baez (May 26 2024 at 22:14):

What's a factorization system for a group, or a classical group?

Notification Bot (May 26 2024 at 22:28):

Cole Comfort has marked this topic as resolved.

Notification Bot (May 27 2024 at 09:24):

Morgan Rogers (he/him) has marked this topic as unresolved.

Morgan Rogers (he/him) (May 27 2024 at 09:29):

I don't know why @Cole Comfort marked this topic as resolved. There can be meaningful factorization systems in a group; if $G$ is isomorphic to a semi-direct product of subgroups $H \rtimes K$ then every element of $G$ factorizes as $hk$ with $h \in H$ and $k \in K$. I can't remember how unique this is (maybe only unique up to conjugation). I also would need to think more to figure out whether such a factorization has the categorical properties one would expect.

Graham Manuell (May 27 2024 at 10:31):

Strict factorisation systems on a group should correspond precisely to Zappa-Szép products of groups.

This is because strict factorisation systems are distributive laws in $\mathrm{Span}$ and Zappa-Szép products of monoids are distributive laws in $\mathbf{B}(\mathrm{Set},\times)$.

Graham Manuell (May 27 2024 at 10:35):

Though note that orthogonal factorisation systems on a group are trivial since both classes must always contain all isomorphisms.

Eric M Downes (May 27 2024 at 11:53):

So... "trivial" here reduces to $(H,K\unlhd G)\,\land \,(H\cap K={1})\implies G\cong (G/H\times G/K)$, or are there other consequences?

Morgan Rogers (he/him) (May 27 2024 at 12:09):

which use of "trivial" are you referring to @Eric M Downes ?

Eric M Downes (May 27 2024 at 12:45):

Whatever sense Graham Manuell meant! :) I am guessing that "trivial" means that the unique arrow making the solid diagram commute is the identity functor $\bf 1_G$ so it looks like you end up with just the product diagram relating $G$ and its quotients, and indeed that coincides exactly with Szep products, though I was unfamiliar with that term. But I'm still learning about factorization systems so that's also kind of a guess.

Morgan Rogers (he/him) (May 27 2024 at 12:54):

A factorization system is trivial if one side or other consists of all morphisms of the category. Since an OFS necessarily contains all isomorphisms in both classes, any OFS is trivial on a group :)

Eric M Downes (May 27 2024 at 12:54):

Ahhh so you're saying its just a repackaging of what we already know about a group.

John Baez (May 27 2024 at 13:51):

By the way, I find "Zappa-Szép product" to be a scary and unenlightening name for this situation: we have a group $G$ with subgroups $H, K \subseteq G$ such that each element $g \in G$ can be uniquely written as $g = h k$ with $h \in H, k \in K$.

Jean-Baptiste Vienney (May 27 2024 at 13:57):

What is more scary is the definition of the external version which gives the conditions in which you can build such a group $G$ from two groups $H,K$.

It is much more complicated than both the external direct product and the external semi-direct product.

Todd Trimble (May 27 2024 at 14:00):

I find Frank Zappa to be a scary musician.

John Baez (May 27 2024 at 14:04):

Jean-Baptiste Vienney said:

What is more scary is the definition of the external version which gives the conditions in which you can build such a group $G$ from two groups $H,K$.

It is much more complicated than both the external direct product and the external semi-direct product.

Yes! But the nice thing is that you don't need to look this up or remember it: you can figure it out starting from the simple description I gave. You say the group $G$ has the cartesian product $H \times K$ as its underlying set, and you figure out what a general multiplication rule $G \times G \to G$ must look like in these terms of operations on $H$ and $K$, given that $H \times \{1\} \subseteq G$ and $\{1\} \times K \subseteq G$ are subgroups.

Jean-Baptiste Vienney (May 27 2024 at 14:11):

Oh, that’s really cool. I’ll try if I can figure out the external version by myself like this.

Jean-Baptiste Vienney (May 27 2024 at 14:16):

And is it really the same as a factorization system if you consider $G$ as a one object category?

Jean-Baptiste Vienney (May 27 2024 at 14:18):

Ok, I don’t understand the details but I think it was explained before.

Jean-Baptiste Vienney (May 27 2024 at 14:37):

Graham Manuell said:

Strict factorisation systems on a group should correspond precisely to Zappa-Szép products of groups.

This is because strict factorisation systems are distributive laws in $\mathrm{Span}$ and Zappa-Szép products of monoids are distributive laws in $\mathbf{B}(\mathrm{Set},\times)$.

Why should your two classes of morphisms be subgroups? Isn't a strict factorization system for a group rather a decomposition of the group in a Zappa-Szép product of submonoids?

Jean-Baptiste Vienney (May 27 2024 at 14:38):

Maybe that's already what you meant. I'm not sure.

Graham Manuell (May 27 2024 at 15:53):

To be honest I didn't really consider that, but I think what I said was right: say $x \in M$. Then $x^{-1}$ factors as $me$. So $1 = x x^{-1} = (xm)e$. But $1 = 1 \cdot 1$ and factorisations are unique. Hence $e = 1$ and $x^{-1}$ lies in $M$. So $M$ is closed under inverses. The same holds for $E$ by duality.

Jean-Baptiste Vienney (May 27 2024 at 16:02):

Graham Manuell said:

To be honest I didn't really consider that, but I think what I said was right: say $x \in M$. Then $x^{-1}$ factors as $me$. So $1 = x x^{-1} = (xm)e$. But $1 = 1 \cdot 1$ and factorisations are unique. Hence $e = 1$

I'm ok up to there. We also have $xm=1$. But I'm not sure whether it implies that $m$ is an inverse of $x$?

Jean-Baptiste Vienney (May 27 2024 at 16:05):

My bad, ok. You just multiply by $x^{-1}$.