# User talk:Lord Farin/Archive 2

*This page contains archived discussions, for the contributor Lord_Farin felt they could be historically - to the extent this term applies to ProofWiki - significant. It is not intended, but may be merited, that these discussions are continued.*

## Orderings on products

I'd like to merge the current Definition:Ordered Product and Definition:Lexicographic Order, extend them to well-ordered index sets, and rename them Definition:Lexicographic Ordering. What's the right way to do such while preserving history, avoiding confusion, etc.? I think Definition:Ordered Product should probably become something of a disambiguation page, pointing to Product Order and Lexicographic Order. --Dfeuer (talk) 21:58, 18 December 2012 (UTC)

- These will be edits that hit a substantial amount of PW. Therefore, it's probably best to first set up the stuff in e.g. your sandbox area. That way, we can tweak and adjust all we want without affecting the main wiki with immature or incomplete material. I've done this in the past, can't remember what section of the site it was, but it worked quite well. --Lord_Farin (talk) 22:13, 18 December 2012 (UTC)

- Found something thornier: the site already has a somewhat different notion of lexicographic ordering, which is not on products at all but essentially on strings whose letters are drawn from a single totally ordered set. So I guess we need two different kinds of lexicographic orderings? The kind on strings appears to be isomorphic to a special case of the one on products, since all that's needed is a product using the naturals as the index set, where the totally ordered underlying set is augmented with a new least element, or so I figure. Dfeuer (talk) 01:09, 19 December 2012 (UTC)

- The one on strings indeed appears to be a special case (similar to $\R^n$ being a special case of Cartesian product). The two are conceptually sufficiently distinct that I'd advise for them to be separate sections (a set-up like Definition:Continuous Mapping, where the real and topological versions are both mentioned). --Lord_Farin (talk) 08:52, 19 December 2012 (UTC)

Can you take a look at User:Dfeuer/Definition:Lexicographic Ordering on Product and the pages it links to? I think they're a decent start on the lexicographic side of things. --Dfeuer (talk) 07:47, 21 December 2012 (UTC)

- Definition appears to be correct information-wise, and is nice, very general :). However, it's not up to house style (yet). Do you want me to fix that? --Lord_Farin (talk) 15:27, 21 December 2012 (UTC)

- You're more than welcome to. Small annoyance: there's a page stating (and beginning to prove) the theorem that the lexicographic ordering of the set of all finite sequences on a well-ordered set with at least two elements is
*not*a well-ordering. A similar result holds for lexicographic orderings on products of infinitely many well-ordered sets, each containing at least two elements, and the proof is essentially identical.

- You're more than welcome to. Small annoyance: there's a page stating (and beginning to prove) the theorem that the lexicographic ordering of the set of all finite sequences on a well-ordered set with at least two elements is
- However, because encoding the finite sequence set as a subset of the product expands each well-ordered set to
- at least three elements, neither theorem seems to imply the other. Can you think of any nice way to hit both at once, or do we need to keep them separate? --Dfeuer (talk) 16:18, 21 December 2012 (UTC)

- I think they'd best remain separate. It would be a good idea (and probably not too hard) to try and give a proof of the finite-sequence version with the more general version. --Lord_Farin (talk) 17:37, 21 December 2012 (UTC)

## Notations change

Sorry for delaying my answer—I fully understand that you could perceive it as ignoring you—I just wanted to make my edits concise. Of course no hard feelings for your blocking action: all in all my fault as, as you noticed, I'm not a part of "main contributing force"; just thought that change of "subgroup product" into "subset product" name won't make any objections… Here I'd like excuse me once more for my hastiness!

As for giving opinions I leaved some in Definition talk:Internal Group Direct Product and had an answer just recently (which I gladly appreciate) but with no rationale for such statement of the definiton… It might be clear to you for you're part of "maintenance team" but it's quite puzzling for me who is outside this group. I'll try to be more careful when making ammendments (not corrections) in the articles and take into heart your pleas but I'd like to ask you for giving more details about the reasons about solutions you percieve as best. Thanks in advance! joel ^{talk} 23:54, 9 January 2013 (UTC)

P.S. I hope that you accept changes I've made as for subset/subgroup product and that they are not a nuissance for the editiorial team…

- For the purpose of deciding notation and nomenclature, it's probably most efficient to call just me an prime.mover the "main contributing force" - notwithstanding the fact that there has been an increase in users which contribute regularly. Maybe I should've stuck with "maintenance team". The edits pertaining to subset product were correct IMO, no worries on that part.
- Usually I try to justify particularly any negative calls I make (not speaking for prime.mover here, but take into account that he has led PW through the dark years when self-proclaimed gods would come in and change notation because they didn't like it, and that such has carved certain instincts in him). Recently, we've been increasingly proactive in this regard. Note e.g. the advent of Help:FAQ which should relieve some of the burden of answering the same questions over and over again.
- In this light the Definition:Internal Group Direct Product page should not be taken as representative. Were such ever to occur again, don't hesitate to kindly ask for an explanation if you feel one is appropriate.
- Finally, on a side note, it's usually a good idea to reply to a comment on the same page (also for discussions in user talk). This ensures comprehensibility and is most likely to get all those interested notified of the reply (via the watchlist updates - users could be watching your, but not my talk and thus miss your reply). --Lord_Farin (talk) 08:27, 10 January 2013 (UTC)

## Random group theory question

Planetmath claims without proof that a totally ordered group with the order topology is a topological group. I'm not really seeing that, though I could of course be missing something. However, is there a term fora group in which every element has a square root? That sort of ordered group would form a topological group for sure. --Dfeuer (talk) 08:14, 28 January 2013 (UTC)

- It appears to me that any subbasis element of the order topology is sent to another subbasis element under left and right multiplication, and also under inversion. That'd make it a topological group in my book.

- I don't know a term for such a group. Perhaps the above comment renders your question obsolete. --Lord_Farin (talk) 09:40, 28 January 2013 (UTC)

- If by a "square root" you mean an element $y$ such that $y \circ y = x$, then this rings a bell somewhere in the back of my mind, but a cursory search has turned up nothing. $C_3$ is such a group, but (unless I misunderstand what I'm talking about) the Klein-4 group is not. Interesting. --prime mover (talk) 09:49, 28 January 2013 (UTC)

- I thought about $(\Q, +)$ and $(\Q_{>0},\times)$ (also for $\R$ and additive $\C$, as well as vector spaces over the fields among these). --Lord_Farin (talk) 09:52, 28 January 2013 (UTC)

- The multiplicative rationals don't have square roots. As for your comment, group multiplication does always send subbasic elements to subbasic elements, but I don't think the inverse images of subbasic elements are necessarily open, unless of course you can explain why they would be. The usual proof that they are (in, say, the reals), relies on being able to take half of something (i.e., a "square root" in a group). And a topological group doesn't require multiplication to be open but rather continuous. --Dfeuer (talk) 14:38, 28 January 2013 (UTC)

- While realising that my earlier arguments didn't work, I think the result can be proved along the following lines (denote $\mu$ for group operation, $\uparrow$ for strict upper closure (lower closure by duality), and assume for convenience the group is abelian):

- $(x,y) \in \mu^{-1}(\uparrow z)$. Suppose $\exists y': y \succ y'\succ z \circ x^{-1}$. Then it can be shown that $(x,y) \in \uparrow(z\circ y'^{-1})\times \uparrow(y') \subseteq \mu^{-1}(\uparrow z)$. In the other case, we have $\uparrow (z\circ x^{-1}) = \bar\uparrow y$ by total ordering, where $\bar\uparrow$ is weak upper closure. In that case, $(x,y) \in \uparrow(z \circ y^{-1})\times \uparrow (z\circ x^{-1})$ and also, that set is in $\mu^{-1}(\uparrow z)$ (here we use the set equality established). Done.

- A proofread on that would be appreciated. --Lord_Farin (talk) 15:22, 28 January 2013 (UTC)

Working on it, but the notation is spinning my head a bit. --Dfeuer (talk) 15:33, 28 January 2013 (UTC)

- $\mu^{-1}(\uparrow z) = \{(x,y) \in G\times G: x \circ y \succ z\}$, $\uparrow z = \{x \in G: x \succ z\}$, $\bar \uparrow z = \{x \in G: x \succeq z\}$. Hope that helps. --Lord_Farin (talk) 15:39, 28 January 2013 (UTC)

I put it in the sand at User:Dfeuer/Totally Ordered Group with Order Topology is Topological Group to see if we can flesh it out to a full proof. --Dfeuer (talk) 21:29, 28 January 2013 (UTC)

## Massive transitive closure rewrite

I think it generally works better now, but:

- The transclusions are kind of a mess.
- The proof of Equivalence of Definitions of Transitive Closure (Relation Theory)/Union of Compositions is Smallest seems a tad inelegant. I wonder if there's some way to directly derive that from Recursive Construction of Transitive Closure.
- Subpage names are not reflected in link names.
- I'm sure I'm missing some issues.

--Dfeuer (talk) 21:14, 2 March 2013 (UTC)

- Definition:Transitive Closure (Relation Theory) looks good now. General desired form of equivalence proofs is as on Equivalence of Definitions of Connected Topological Space. More comments will appear here as I progress. — Lord_Farin (talk) 22:26, 2 March 2013 (UTC)

- Header issue at equivalence proof tackled; the extension I wrote precisely serves this purpose. Although it is currently bugged (TOC doesn't work properly, and same will hold for section editing (but the latter is discouraged anyway) - rest assured, it is being worked on). — Lord_Farin (talk) 22:49, 2 March 2013 (UTC)

- Maybe I'm able to assess that once the structure of Equivalence of Definitions of Connected Topological Space is into place (and that's a genuine comment, not a poke to remind that such still has to be done). — Lord_Farin (talk) 23:29, 2 March 2013 (UTC)

- I think I see what you mean now. Done. I'm reconsidering the numbering, though. The definitions that strike me as the most useful for most purposes are the definition as the smallest transitive superset and the one I awkwardly/imprecisely called "finite chain". Of course, I'm no grand master of relation theory, so I could be wrong about that. --Dfeuer (talk) 00:13, 3 March 2013 (UTC)

- Numbering is in some sense arbitrary. I no longer consider it very interesting — let alone worth the hassle of moving pages in two stages just to juggle them around (to which effect possibly the descriptive subpage names may be an aid). Off to bed now, with the one suggestion that each of the section titles gets a number indication of what it proves (e.g. $(1) \iff (4)$). — Lord_Farin (talk) 00:19, 3 March 2013 (UTC)

- Oh, and good job :). — Lord_Farin (talk) 00:19, 3 March 2013 (UTC)

## Topological closures?

You wanted to do something with them? When you do, please see what can be done about the rather awkward names/lack of transclusion/whatever for Closure is Closed/Closure Operator, Intersection of Closed Sets is Closed/Closure Operator, and Set Closure is Smallest Closed Set/Closure Operator. Also, any fresh approaches to any proofs in the newly-created Category:Closure Operators would be appreciated. --Dfeuer (talk) 00:04, 4 March 2013 (UTC)

## But I don't know if they're already here, what names they go by, etc.

But I don't know if they're already here, what names they go by, etc.

### 1

If an algebraic system is closed, then it is also closed under all finite "compositions" of its operations. Simple example: since the reals are closed under addition and multiplication, they're also closed under the $4$-ary operation mapping $(a,b,c,d)$ to $(a+bc)d$.

- This is interesting and intuitive. I don't recall ever having seen it (it's too obvious to be in many texts, presumably). Suggest searching for it. It's definitely worth having but I'd rather use the literature terminology (if it exists). — Lord_Farin (talk) 09:00, 9 March 2013 (UTC)

### 2

Going the other way, if whenever the results of certain operations have a certain property, then their operands do, then this holds also for operations built from them. For instance, a composition of two idempotent inflationary mappings having $x$ as a fixed point means each of them does. Thus if $x$ is a fixed point of $f \circ g \circ h$ it is a fixed point of each. --Dfeuer (talk) 23:27, 8 March 2013 (UTC)

- I fail to see how you generalise from that instance to your first sentence. Perhaps you could give another example? — Lord_Farin (talk) 09:00, 9 March 2013 (UTC)

- These results don't currently exist on PW but there's nothing to stop you adding them. If you can prove them. The understanding is that this particular venture is your own project, mind, you can't just post other people and ask them to finish it off for you. --prime mover (talk) 09:07, 9 March 2013 (UTC)

- PM:I was looking for help finding the appropriate names (and theorems, if they were already up). LF: It's really the same thing run in reverse. Another way of looking at the same theorem, I now realize. In the strictly positive
~~reals~~integers, if $ab+c \preceq 5$ then $a,b,c \preceq 5$--Dfeuer (talk) 15:23, 9 March 2013 (UTC)

- PM:I was looking for help finding the appropriate names (and theorems, if they were already up). LF: It's really the same thing run in reverse. Another way of looking at the same theorem, I now realize. In the strictly positive

- And I said, the results don't exist on PW. As for a name, "generalised closedness theorem" might be adequate. I was referring to a propensity to start something, lose direction, and then need to ask for help. This can distract contributors from their own projects. --prime mover (talk) 15:58, 9 March 2013 (UTC)

- Now I understand - thanks. I've been contemplating 1, and it seems that what is being asked here is related to logic, namely the generation of terms in formal languages does something like this. It's kind of a generalization to composition to higher arity. If I have some more time I might crawl through some source works to see if they've got a word for it - probably not, in linear texts these kinds of things usually are left implicit or are not named but simply mentioned in the prose. — Lord_Farin (talk) 15:42, 9 March 2013 (UTC)

## Terminology clash

Smullyan and Fitting use the term nest to mean a class of sets which is totally ordered by inclusion. No problem. Where things get dicy is that they define a chain to be a nest which is also a set. This clashes with the order-theoretic notion of a chain. How would you suggest this conflict be resolved? --Dfeuer (talk) 20:55, 25 March 2013 (UTC)

- This restriction on the term "chain" appears to me as idiosyncratic. Suggest to add a note to the "Also defined as" section of chain but keeping def the same. Then where Smullyan/Fitting write "chain" we can write e.g. "chain of subsets" or something like that if it is to be stressed that we consider $\subseteq$. I don't deem it worth the hassle of putting up a separate case. — Lord_Farin (talk) 22:06, 25 March 2013 (UTC)

- Both "Nest" and "Chain" seem appropriate, although perhaps not conveying everything. Other than that, I only mention that a proper way of distinguishing sets from classes would be a non-trivial part of a paradigm for set theory. — Lord_Farin (talk) 22:23, 25 March 2013 (UTC)

## Dimension theorem help

My proof of the Dimension Theorem for Vector Spaces has a gap. I believe it can be filled by this lemma, but I haven't been able to figure out how to prove it (assuming it's true). It's essentially a stronger form of Size of Linearly Independent Subset is at Most Size of Finite Generator:

Let $B$ be a basis of a vector space $V$.

Let $L$ be a finite linearly independent subset of $V$.

For each $x \in V$, $x$ can be expressed uniquely as a linear combination of elements of $B$. (Expression of Vector as Linear Combination from Basis is Unique only shows this for finite $B$, but we should be able to fix that, I think).

Let $\Phi(x)$ be the vectors in the expression of $x$ (with non-zero coefficients).

Then: there is a injection $f: L \to B$ such that $f(x) \in \Phi(x)$ for each $x$.

We know from another theorem that for each $F \subseteq L$, there is an injection from $F$ into $\bigcup \Phi(F)$, and I have the feeling that may be enough to lead to some sort of combinatorial proof (with no further need for any information about vector spaces), but I don't know. The situation can get a bit tricksy:

If $\Phi(x) = \{a, b\}$, $\Phi(y) = \{b,c\}$, and $\Phi(z) = \{a,c\}$, for example, then if you set $f(x) = a$ and $f(y) = c$, then you won't be able to choose a value for $f(z)$.

However, you can set $f(x) = a$, $f(y) = b$, and $f(z) = c$ just fine.

Of course, you could have examples where only some of the $\Phi$ values intersect, etc. Any ideas? --Dfeuer (talk) 22:34, 21 May 2013 (UTC)

My conjectural lemma to the conjectural lemma is at User:Dfeuer/Condition for Injective Choice Function.

- OK, so I've found out that the general version of this lemma is called Hall's Marriage Theorem. Wikipedia has what claims to be a proof of the finite form in the language of graph theory, which is entirely incomprehensible to me at this juncture. But at least I know it's true. --Dfeuer (talk) 01:46, 22 May 2013 (UTC)

- Seems like you've taken yourself a nice part of the way already. For the finite case (which seems to be what you're dealing with here anyway) you could perhaps take -- for bases $B, B'$ and $\Phi: B' \to B$ -- the poset $\{F \subseteq B': \text{there is an injective choice function for $\Phi\restriction_F$}\}$ by inclusion, which is finite, hence has upper bounds for chains, and then apply Zorn in the standard way to obtain $B'$ as the only maximal element. I don't know if it follows through, though. — Lord_Farin (talk) 07:58, 22 May 2013 (UTC)

- Back up. We don't need Zorn by any means. I believe the dimension theorem is a fairly simple consequence of Hall's Marriage Theorem (for infinite sets of finite sets), which I believe is a straightforward application of the Cowen-Engeler Lemma
*once the finite case is proven*. I've found a few proofs of the finite case. I have not yet, however, found one I actually understand properly. They vary from vague descriptions of algorithms to inscrutable (to me) applications of graph-theoretical concepts. --Dfeuer (talk) 09:18, 22 May 2013 (UTC)

- Back up. We don't need Zorn by any means. I believe the dimension theorem is a fairly simple consequence of Hall's Marriage Theorem (for infinite sets of finite sets), which I believe is a straightforward application of the Cowen-Engeler Lemma

- OK, I think I've gotten most of the proof chain together now: Hall's Marriage Theorem/Finite Set, Hall's Marriage Theorem/General Set, and Dimension Theorem for Vector Spaces. It's kind of ragged, though. Do you see any obvious ways to improve things (aside from adding necessary links)? --Dfeuer (talk) 01:15, 24 May 2013 (UTC)

- Line of thought seems good. I have reworded a tiny bit in the dimension theorem, otherwise I see no obvious way to improve things (which does not imply such does not exist, of course). — Lord_Farin (talk) 07:34, 24 May 2013 (UTC)

## Absolute convergence stuff

I added a couple thoughts to Talk:Absolutely_Convergent_Generalized_Sum_Converges. We don't currently have definitions here for normed semigroup or normed group, and I don't have appropriate sources, but the concept is trivial: A normed semigroup is a semigroup with a norm satisfying the triangle inequality. A normed group additionally obeys $\lVert g \rVert = 0$ iff $g$ is the identity and $\lVert g^{-1} \rVert = \lVert g \rVert$. A normed semigroup seems to be the most general context for "absolutely convergent infinite series", a normed commutative semigroup seems to be the most general context for "absolutely convergent generalized sum", and a complete normed abelian group seems to be the most general context in which absolute convergence implies convergence. Requiring a Banach space seems like overkill if the underlying field/division ring never even appears! --Dfeuer (talk) 21:00, 1 August 2013 (UTC)

- While you are right, there's not much added value if the definitions you mention aren't found in the literature. Maybe in five years, when I get back to the functional analysis department (it's going to happen
*some*day, I'm sure. Just be patient :)). — Lord_Farin (talk) 21:04, 1 August 2013 (UTC)

- My usual standing order: do not just enter random stuff that you think you know, particularly definitions and refinements of existing definitions. If the context is not in place it reduces the comprehension value of the existing material. --prime mover (talk) 21:13, 1 August 2013 (UTC)

I just put up a somewhat rough Positive-Term Generalized Sum Converges iff Supremum. This leads immediately to the conclusion that generalized sums are monotone, in a stronger sense than the theorem you put up by that name. Specifically, the generalized sum over any subset of the index set, not just finite ones, is less than or equal to the total sum, and this holds for complete ordered abelian groups in general, not just the reals. --Dfeuer (talk) 00:24, 2 August 2013 (UTC)

- I just did a bunch of paper-and-pencil work on some of this stuff on the plane ride to Denver, CO (Hurray! Vacation!). It looks like the progression should begin like this:

- 1. Let $(L,<,\tau)$ be a LOTS. Then an increasing net in $L$ converges to a point iff that point is the supremum of its image. If the order is Dedekind-complete, an increasing net converges iff it is bounded.
- 2. Let $(G,\cdot,\le)$ be a totally ordered abelian group, considered under the order topology. Let $\{x_i\mid i\in I\}$ be an indexed subset of (weakly) positive elements of $G$. Then the net mapping each finite subset of $I$ to the sum of the $x_i$ values is increasing.
- 3. In the situation in (2), a convergent subsum is less than or equal to a convergent sum. If the order is complete, then a subsum of a convergent sum of positive values always converges.
- 4. A sum including both positive and negative values converges iff both the positive and the negative values have convergent sums, which happens iff the sum is absolutely convergent (absolute value for an
*ordered group*—not a norm). - 5. If we end up with a first-countable topology and an Archimedean totally ordered abelian group, I believe that's enough to show that a generalized sum can only converge if all but countably many terms are $0$. Archimedean, if I'm remembering correctly, implies that for each $x>0$, we can only have finitely many terms at least as big as $x$. First-countable, I believe, shows that a sequence can get us close enough to $0$, and we can take a countable union of finite sets to capture all the positive terms. I still need to finish working this step out, and there may be a better way to deal with it, but it certainly works out very easily for $\Bbb R$. --Dfeuer (talk) 21:51, 2 August 2013 (UTC)

- Interesting, and nice work. I would enjoy seeing it fleshed out (but preferably in your own zone until we find some sources dealing with this mostly functional-analytic construct "generalized sum" in the distinctly more abstract situation of a totally ordered abelian group. — Lord_Farin (talk) 07:23, 3 August 2013 (UTC)

## Definition:Boolean Algebra/Definition 3

Hope this doesn't throw you off course too badly ... what's your thoughts on removing this definition? It very much duplicates Definition 1 except that it adds the complementation operator during the exposition of the axioms rather than as one of the elements of the structure. Df and I briefly discussed this back in April. He suggested we wait for you to get back. My money is on deleting it. --prime mover (talk) 06:00, 15 September 2013 (UTC)

- To me, there's an important semantic difference. Institutionalising an operator and deriving its well-definedness are two very different conceptual approaches. It's an "easy to use"-"easy to prove" duality, so to say. I hope to get to the part of model theory that allows for rigorous proofs that adding such extra function symbols does not change the expressivity (or strength, if you wish) of the language. I'll rethink in time, if you're fine with that (some extra time after five months won't be too big an issue). — Lord_Farin (talk) 14:57, 15 September 2013 (UTC)

- No urgency - I just happened to pass by on that side of the road again this weekend, and it occurred to me there was that small piece of unfinished business. --prime mover (talk) 19:05, 15 September 2013 (UTC)

## DeleteRedirect: not sure about it

The problem with the DeleteRedirect template is that it breaks the view of what links via the redirect when using the What Links Here tool.

As an example, the page Euler Phi Function of an Integer/Corollary has Euler Phi Function Even for Argument Greater than 2 linking to it (as it wasn't moved when the page name was changed - my omission).

However, if you do a "What links here" from Euler Phi Function of Integer/Corollary, that link doesn't show up. There's a danger that an incautious editor might just look at Euler Phi Function of Integer/Corollary, see Euler Phi Function of an Integer/Corollary but with nothing linking to it, think "Nothing links to this redirect", forgetting that **it is no longer a redirect** and delete it prematurely.

In a perfect world, when such a page is renamed, one would go through painstakingly replacing all the links (which I try to do, but it cannot be guaranteed that I will) -- but when the list is large, this task may be postponed, and then forgotten about.

Just something to bear in mind ... --prime mover (talk) 11:57, 18 September 2013 (UTC)

- Yes, we have to be more careful when deleting. IMO that is outweighed by the benefit of eventual
**deletion**without losing reliability to external parties, as I set out on Talk:Main Page. — Lord_Farin (talk) 12:01, 18 September 2013 (UTC)

- I can't argue against the fact that it seems like a good idea to do (I've already started work on a few of the deletions) - but I'm worried that we'll end up deleting too much. Also see my comments on the main talk page. --prime mover (talk) 12:11, 18 September 2013 (UTC)

## question about page

What does Mappings to Algebraic Structure form Similar Algebraic Structure mean? It starts with some "Let"s and a "Define" but it doesn't actually draw a conclusion. There's no "then".

Can you revisit this? Thx. --prime mover (talk) 20:48, 7 October 2013 (UTC)

- While I'm about it, same applies to Mappings to R-Algebraic Structure form Similar R-Algebraic Structure. --prime mover (talk) 20:49, 7 October 2013 (UTC)

- Digging deep into my memory, these pages where created from following red links, from Mappings to R-Algebraic Structure form Similar R-Algebraic Structure and Definition:Pointwise Scalar Multiplication of Mappings, respectively.

- The gist is that Definition:Operation Induced on Set of Mappings gives rise to general theorems that e.g. mappings to a group form a group. The purpose of the two pages was to bundle together all those results via transclusion. I hope that makes sense.

- They're very much unfinished. Do with them as you wish. — Lord_Farin (talk) 21:08, 7 October 2013 (UTC)

- Okay no worries, I'll leave it for now. --prime mover (talk) 06:12, 8 October 2013 (UTC)