Definition talk:Complement of Subgroup

Good call; I was not aware of Definition:Independent Subgroups. Though Definition talk:Complement (Group Theory) also requires that every element of $G$ is a product of elements of $H$ and $K$, in addition to $H$ and $K$ being independent. In any case, a link to Definition:Independent Subgroups would do no harm. --barto (talk) 06:59, 11 July 2017 (EDT)

I think it's over-egging it to implement two definitions, because they both say exactly the same thing. The equivalence proof is nothing more and nothing less than that intersection is commutative. Surely we don't need two separate definitions just for that? --prime mover (talk) 07:06, 11 July 2017 (EDT)

There's also the $G=HK$ vs. $G=KH$. --barto (talk) 07:09, 11 July 2017 (EDT)

In general, if $K,H$ are subgroups of $G$, $HK\neq KH$. --barto (talk) 07:10, 11 July 2017 (EDT)

Well yes, but $K \cap H = H \cap K$ trivially. So $K H = K \cap H$ and $H K = H \cap K$ just by renaming the subgroups. Is there a subtlety here that I'm completely missing?

Oh, and as for "Definition talk:Complement (Group Theory) also requires that every element of $G$ is a product of elements of $H$ and $K$, this follows from Internal Direct Product Theorem so one follows from the other, yes? Unless I'm missing something obvious again. It's a long time since I did group theory and I probably misunderstood it completely even then. --prime mover (talk) 07:17, 11 July 2017 (EDT)

$HK$ here means subset product, so $HK\neq H\cap K$. If $H$ and $K$ are normal (which is not required here) then we're in the case of Internal Direct Product Theorem indeed. But what we're really using here is the following fact (worthy of a page on $\mathsf{Pr} \infty \mathsf{fWiki}$): if $H,K\subset G$ are subgroups such that $HK$ is a group, then $HK=KH$. The proof really takes only 1 line, but there's something to prove here. --barto (talk) 07:30, 11 July 2017 (EDT)

This: Subset Product of Subgroups. Maybe it is cleaner to first create a definition page for $G=HK$, but I'm not sure how to refer to that phenomenon. --barto (talk) 07:42, 11 July 2017 (EDT)

Yes I know $KH \ne HK$ in general. That I'm not disputing. That's not the point. My bone of contention is that "$KH = G$ and $K \cap H = \{e\}$" is saying exactly the same thing as "$HK = G$ and $H \cap K = \{e\}$" -- because all you need to do is rename the subgroups. And changing $H \cap K = \{e\}$ for $K \cap H = \{e\}$ is too trivial to mention.) $H$ and $K$ are non-distinguished, they can be any subgroups. So: "$H$ and $K$ are complementary if and only if: 1. their subgroup product is the whole group, and 2. their intersection is the identity." Exchanging $H$ and $K$ does not make a new definition, all you're effectively doing is renaming the subgroups. And yes, in this context, because of Subset Product of Subgroups, $HK$ does always equal $KH$ because $G$ is a subgroup of itself.

The equivalence proof is: "Rename $H$ to $K$ and $K$ to $H$ and note that $H \cap K = K \cap H$ by Intersection is Commutative. I'm still completely missing the subtlety. --prime mover (talk) 08:04, 11 July 2017 (EDT)

I understand it is confusing. At this point it is important to distinguish between

$K$ is a complement of $H$

$H$ is a complement of $K$

And (not yet defined, but you mentioned it)

$K$ and $H$ are complementary

$H$ and $K$ are complementary

With this in mind, we can not just rename things to conclude that the definitions are equivalent. (It is like saying: $H$ is a subgroup of $G$ is equivalent to $G$ being a subgroup of $H$.) To convince you that the linguistic argument is invalid, the definitions are not equivalent in more general setttings (see below), so there is really something special about groups we're using here, namely, the existence of an inverse.

Here's an example of (unitary) monoids where the definitions are not equivalent: Let $G$ be the monoid of words on the alphabet $\{a,b\}$ with at least as many $a$'s as $b$'s (including the empty word). Let $H$ be the submonoid of words of even length starting with $b$ (including the empty word), $K$ the submonoid of words starting with $a$ (including the empty word). Then $HK=G$, but $KH$ does not contain odd words starting with $b$. ($HK=G$ requires some analyzing in the case $g\in G$ starts with $b$ and has odd length.) --barto (talk) 10:02, 11 July 2017 (EDT)

As you say, we're really using Subset Product of Subgroups to show that the definitions are equivalent. --barto (talk) 10:05, 11 July 2017 (EDT)

All you need to do is establish that the the group complement of the group complement is (trivially) itself, and job is done.

What's your source for this? I need to read up on it, because either I've completely missed the point you're trying to make, or you've missed the point I'm trying to make. This is exactly why I mistrust definitions which are not backed up by the literature. --prime mover (talk) 11:46, 11 July 2017 (EDT)

Unfortunately course notes, mostly. You can find this in Dummit & Foote (2003) - Abstract Algebra; §5.5 Semidirect products, the definition under Theorem 12. The counterexample for monoids is one I invented today. --barto (talk) 12:45, 11 July 2017 (EDT)

The best approach here is to set up the page to define the complement for a monoid, then, and write whatever results are needed to explain that specialness of groups. At the moment it just hangs there without a context --prime mover (talk) 05:27, 12 July 2017 (EDT)

I'd like to, but my digression about monoids was just to illustrate the importance of working with groups. I don't know if 'complement' is defined for submonoids, or what that definition would look like. I doubt it, because of the arising ambiguity. --barto (talk) 06:27, 12 July 2017 (EDT)