Smith typeset 16:23 30 Jan 2005 loop inclusions

Inclusions among diassociativity-related loop properties

Warren D. Smith [email protected]

January 30, 2005

Abstract — We attempt to find all implications among Sometimes the loop operation is regarded as multiplication (in 19 commonly used diassociativity, Moufang, Bol, alterna- which case we usually call the identity 1), other times it is re- tivity and inverse-related properties in loops. There are 6 garded as addition (in which case we usually call the identity among these that appear to be valid in finite but not infi- 0). We shall use both notations in this paper. nite loops. Under that assumption, we completely settle Probably the most widely studied properties of loops are: the problem. We study in detail the apparently-simplest : the property of being a group, i.e. of obeying the among the 6 nasty cases: the “LRalt=⇒2SI” question of associative law x · yz = xy · z; whether a left- and right-alternative loop necessarily has Moufang: the Moufang property1 (x · yz)x = xy · zx, equiv- 2-sided inverses. We construct an infinite loop in which alent to obeying both the left-Bol x(y · xz) = (x · yx)z this is false. However, X must have a 2-sided inverse in n and right-Bol x(yz · y) = (xy · z)y properties. any LRalt loop with ≤ 185 elements or in which X = 1 Lalt: the left-alternative law x · xy = xx · y; with n ≤ 13 (M.K.Kinyon has improved “13” to “31”), re- Ralt: the right-alternative law yx · x = y · xx; sults suggesting this is the case in all finite loops. The Flex: the flexible law xy · x = x · yx; problem of fully resolving this may be the hardest natu- LIP: the left-inverse-property (1/x) · xy = y; ral problem in that is this simply posed. RIP: the right-inverse-property yx · (x\1) = y; Antiaut: the law 1/(xy) = (1/y)(1/x) of antiautomorphic inverses; Contents 2SI: the law of 2-sided inverses 1/x = x\1; PA: power-associativity (the statement2 that xn is unam- 1 Introduction 1 biguous for all positive integer n); and 3PA: 3-power-associativity xx · x = x · xx; 2 Which subsets of properties imply which? 2 Diassoc: diassociativity (the statement that any two ele- 3 Collected counterexample loops 3 ments of L generate a subgroup). 4 DoLRaltloopshave2-sidedinverses? 9 Despite the large amount of study devoted to these proper- 4.1 A countably infinite LRalt loop without 2-sided ties, many fundamental questions about them had never been inverses ...... 9 answered. Foremost among these include 4.2 Do finite LRalt loops have 2-sided inverses? . . 9 1. Which subsets of these properties imply which others? 4.3 Thegraphpicture ...... 13 2. Is there a finite equational basis for (finite set of equa- 4.4 Possible87%solution? ...... 13 tions implied by and implying) diassociativity? 4.5 Candidate for the most frustrating problem in The latter question is settled in the companion paper [20]: theworld?...... 14 loop-diassociativity has no finite equational basis. The 5 Acknowledgements and updates 14 present paper attacks the former question. References 15 The attack is initially straightforward: we consider all pos- sible subsets among these properties and decide which ones are achievable. Our achievability proofs are simply specific constructions of finite loops, and our unachievability proofs 1 Introduction are sequences of logical deductions. A magma is a set L equipped with a binary operation ab. A However, a surprising development prevents this attack from is a magma in which there exists a unique solution attaining victory: it appears there are 6 implications among x to yx = z (usually denoted x = y\z) and to xy = z (usu- properties which are true in all finite loops (so that no finite ally denoted x = z/y). A loop is a quasigroup in which there counterexample exists) but false in certain infinite loops (pre- exists an identity element e so ex = xe = x for all x ∈ L. venting any “pure proof” of that implication i.e. via any finite (Colloquially: “a loop is a non-associative group.”) sequence of deductions in first order logic).

1 There are 4 Moufang identities, all equivalent by lemma 3.1 p.115 of [2]. The other three are x(yz · x)x = xy · zx, (xy · z)y = x(y · zy), and y(z · yx) = (yz · y)x. 2Warning: Power-associativity is defined slightly differently in the companion paper [20].

Feb 2004 1 1. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions Define a loop to be LR-alternative if it is both L- and R- Then our loop properties obey the inclusions in figure 2.1. alternative, IPLR if it is both LR-alternative and IP, alter- All the inclusion relations in figure 2.1 are well known and/or 3 native (Alt) if it is both LR-alternative and flexible, and easy except for theorem 1 and these three IP-alternative if it is both IP and alternative, i.e. both IPLR and flexible. 1. Bol loops are power-associative [18]. Consider the implications in loops in table 1.1. I do not believe 2. Moufang loops are diassociative. This is “Moufang’s these are the only 6 implications of this finiteness-dependent theorem” of 1933. Section VII.4 page 117 of [2] proves kind in loop theory; instead I suspect that the world of loops the stronger statement that in a Moufang loop, if ab·c = is absolutely rife with them. a · bc then a,b,c generate a subgroup. 3. The question of whether LR-alternative loops have 2- # implication n sided inverses (shown with dashed line in figure) turns 1 LRalt =⇒ 2-sided inverses 185 out to be remarkably complicated, and will be discussed 2 Flexible ∧ Ralt ∧ LIP =⇒ Lalt 38 later. 3 Flexible ∧ Ralt ∧ LIP =⇒ RIP 36 4 Alt ∧ LIP =⇒ IP * Groups 5 Alt ∧ antiaut =⇒ IP 19 6 Lalt ∧ Ralt ∧ RIP =⇒ IP 17 R-Bol Moufang loops

Figure 1.1. 6 implications conjectured to be true in finite R-alt RIP diassociative L-Bol R-Bol but false in infinite loops. Each of the implications is true in all loops with ≤ n elements for the value of n tabulated IP-alternative Power-associative mace4 (proven by exhaustive search using [10]). In §4.1 we show statement 1 is false in an infinite loop, so that Inverse IPLR alternative no “pure” proof of it can exist. Searches with otter [11] show property N there are no short pure proofs of statements 2-6. LR-alternative Here is my effort to find the simplest example of a finiteness- LIP RIP flexible dependent fact about loops: L-alt R-alt antiaut Theorem 1 (PA=⇒2SI). implies 2- F sided inverses in finite loops, but not in infinite loops. 3PA n−1 Proof: An element X in a finite loop obeys XXℓ = 1 for n−1 some n ≥ 0, so by power-associativity Xℓ X = 1 proving X loops with two-sided-inverses −1 n−1 has a 2-sided inverse X = Xℓ . (For exponent notation and the fact n exists see EQ 12 and lemma 4.) But the infinite Figure 2.1. Taxonomy of loop-type inclusions. (A much LRalt loop we shall construct in §4.1 is power associative but larger version of this taxonomy will be in the upcoming book  lacks 2-sided inverses. [21].) N Obviously, if one of the implications in table 1.1 is false in some infinite loop, then there cannot be a pure proof of it. It In the presence of antiaut, any left-property and its mirror is less obvious that the reverse is also true: right-property imply one another, e.g. antiaut causes 2-sided inverses and causes Lalt to imply Ralt. Also, of course, any Theorem 2. If any of the 6 implications in table 1.1 has no logical statement (such as R-Bol=⇒RIP, proven by Bol [1]) pure proof, then there is a countably-infinite (or finite) loop always has exactly the same validity as its mirrored version in which that implication is violated. (in this case L-Bol=⇒LIP). Proof: Follows immediately from “G¨odel’s completeness the- Here are statements and proofs of several implications: orem for first-order logics” [4][5][7][14].  1. L-Bol∧Flexible=⇒Moufang. Proof: Simply apply the flexible identity to the term in parentheses on the right 2 Which subsets of properties imply hand side of the L-Bol identity to get the last Moufang which? identity from footnote 1. 2. L-Bol∧Ralt=⇒Moufang. Proof: Rename yx to be Q in Any two among {LIP, RIP, antiaut} implies the third4. A the LBol identity (x · yx)z = x(y · xz) to get xQ · z = loop with these three properties is said to have the “inverse x((Q/x) · xz). Now let y = Q/x to get the Moufang property” (IP).5 identity (x · yx)z = x(y · xz). 3Some other authors have used “alternative” to mean what we call “LR-alternative.” 4See EQ 1.4-1.8 page 111 of [2]. 5We also mention Osborn’s [15] “Weak Inverse Property” y((xy)\1) = x\1. We have not seen these two remarks previously: WIP together with any one among {LIP, RIP, antiaut} suffice to imply the full inverse property IP. Also WIP and Lalt together imply that a loop is Ralt. Another candidate for an implication true in finite but not in infinite loops is that WIP and Lalt together imply IP. This is true in loops with ≤ 11 elements. It appears that our flagship question of whether LRalt−→2SI is unaffected by also assuming WIP and the automorphic inverse property AI. Exhaustive search shows that every WIP and LRAlt loop with ≤ 45 elements has 2-sided inverses, but otter indicates that there is no short pure proof of that, and the infinite loop in §4.1 obeys both WIP and AI.

Feb 2004 2 2. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions 3. L-Bol∧RIP=⇒Moufang. Because L-Bol=⇒LIP and It then becomes a matter of working through the 64 possibili- LIP∧RIP=⇒IP=⇒antiaut, and antiaut converts L-Bol ties with the help of mace4 and (my own program) loopbeaut. into R-Bol. In all 64 cases either mace4 was able to create an example 4. LIP=⇒2SI. Proof: the LIP with y = 1/x gives (1/x) · loop, or such an example arises as a direct product of two x(1/x)=1/x; and since xy = x =⇒ y = 1 we have mace4 discoveries. The examples are compiled in §3. Hence: x(1/x)=1. 5. Lalt∧WIP=⇒Ralt: Theorem 3 (Main result). Under the assumption that the To prove: a loop obeying Lalt xx · y = x · xy and WIP 6 implications in table 1.1 hold in finite loops, figure 2.1 x((yx)\1) = y\1 must obey Ralt xy · y = x · yy. (i) From Lalt and the definition of \ we find x · 1. lists all inclusion-relations among the 19 finite-loop x((xx)\y) = y, x((xx)\y) = x\y, and (xx)\y = properties therein; x\(x\y). 2. all those inclusions are strict; 19 (ii) From Lalt and the definition of / we find 3. all 2 possible subsets of these properties are achieve- (y(yx))/x = yy, then by replacing x with y\x we able except for those forbidden by the implications listed get (yx)/(y\x) = yy, then by taking x = 1 we get throughout the text of this section. In other words, those y/(y\1) = yy. implications are the full set; there are no others. (iii) From the final identity in i using the above expres- Additional kinds of loops will be permitted if any of the im- sion for xx we get (x/(x\1))\y = x\(x\y). plications in table 1.1 are invalid. (iv) From the definitions of / and \ we have: (y/x)\y = x and y/(x\y)= x. and then by taking y = 1 in these we have: 3 Collected counterexample loops (1/x)\1= x and 1/(x\1) = x. (v) From WIP and \ we have (xy)\1 = (y\(x\1)) All have been “beautified,” i.e. their elements have been rear- from which using the final identity in iv we deduce ranged and renamed in an effort to make the loop’s structure xy =1/(y\(x\1)). maximally apparent from its table. Most are minimum pos- Finale: if Ralt were untrue, i.e. A and B existed sible cardinality. In all cases the identity element is e = 0. so that (AB)B 6= A · B, then from Lalt and the ex- “Mirror” examples with all left-handed properties changed to pression for BB in ii we would conclude (AB)B 6= right-handed ones and vice versa, may be got by transposing A · B/(B\1), and then by combining this with the con- the matrix and hence are omitted. Taking the direct prod- clusions of ii, iii, iv, v we could derive the contradiction: uct of two loops intersects their property-sets. This trick is 1/(B\(B\(A\1)))\ 6=1/(B\(B\(A\1))). QED. very useful both for reducing the number of counterexam- ples needed, and for constructing counterexamples too large The fact that there are no other inclusion relations besides for brute force computer searching to find. Although we un- the ones in the figure is proven by constructing counterexam- doubtably could have used products more, we have chosen to ple loops. (For example, the are Moufang but not present non-product constructions whenever small ones are a group.) The ones we tabulate throughout section 3 more available. than suffice for that purpose except that there are two in- stances where we were unsuccessful at constructing either a * 01234 finite counterexample or an inclusion proof. These two cases 0 01234 are shown with dashed lines in figure 2.1: the LRalt=⇒2SI 1 12403 problem and PA=⇒2SI (settled in theorem 1). 2 23140 How can we attack the question of which subsets among the 3 34021 19 properties in figure 2.1 imply which? An equivalent ques- 4 40312 tion is: which of the 219 = 524288 possible property-subsets are achievable in loops? Figure 3.1. 5-element loop. Not PA3, 2SI. N Upon requiring the property subset to obey the inclusions indicated by both the undashed lines in figure 2.1 and PA=⇒2SI, the number of possibilities shrinks6 to 324. If * 0123456 we then also use the fact that antiaut causes any property to 0 0123456 imply its mirror property, it shrinks to 202. If we then also 1 1234560 employ the implications that any two among {LIP, RIP, anti- 2 2361045 aut} implies IP, and that LRalt=Lalt∧Ralt, Alt=LRalt∧flex, 3 3015624 IPalt=alt∧IP=IPLR∧flex, IPLR=IP∧LRalt, and 4 4506213 Moufang=L-Bol∧Ralt=L-Bol∧Flex=L-Bol∧RIP, it shrinks 5 5642301 to 79. Further adjoining all 6 of the implications in table 1.1 6 6450132 would shrink the count to 64. Actually, because some of the 64 sets are there twice (in mirror-duplicated form) there are Figure 3.2. 7-element loop. PA3, but not LALT, RALT, really fewer to worry about. 2SI. N 6Shown by exhaustive computer checking of the original 219.

Feb 2004 3 3. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions * 012345 Figure 3.8. 7-element loop. PA3, 2SI, xx = e, but not PA, 0 012345 LIP, RIP, LALT, RALT, FLEX, Antiaut.N 1 123450 2 234501 * 012345 3 305214 0 012345 4 450123 1 103452 5 541032 2 230514 Figure 3.3. 6-element loop. LALT, but not RALT, 2SI.N 3 354021 4 425103 * 012345 5 541230 0 012345 1 123054 Figure 3.9. 6-element loop. PA, xx = e, but not LIP, RIP, 2 250413 LALT, RALT, FLEX, Antiaut.N 3 304521 4 435102 * 01234567 5 541230 0 01234567 1 12345670 Figure 3.4. 6-element loop. 2SI, but not LIP, RIP, Antiaut, 2 23576014 PA3.N 3 35461702 * 012345 4 47650321 0 012345 5 56017243 1 123450 6 64702135 2 240513 7 70123456 3 354021 4 435102 Figure 3.10. 8-element loop. LIP, PA3, but not PA, RIP, 5 501234 LALT, RALT, FLEX, Antiaut.N

Figure 3.5. 6-element loop. LIP, but not RIP, Antiaut, * 0123456 PA3.N 0 0123456 * 012345 1 1204563 0 012345 2 2016345 1 123450 3 3465201 2 240513 4 4531620 3 351024 5 5640132 4 435201 6 6352014 5 504132 Figure 3.11. 7-element loop. PA, LIP, but not RIP, LALT, Figure 3.6. 6-element loop. Antiaut, but not LIP, RIP, RALT, FLEX, Antiaut.N PA3.N Construction 3.12. LALT, 2SI, but not PA, LIP, RIP, RALT, * 01234567 FLEX, Antiaut: Get a (12 · 21 = 252)-element example by 0 01234567 taking direct product of 3.13 with 3.35. 1 12345670 2 25476301 * 0123456789AB 3 36507214 0 0123456789AB 4 43610725 1 123056B89A74 5 54761032 2 23016B49A785 6 67052143 3 30127896B45A 7 70123456 4 456789AB0123 Figure 3.7. 8-element loop. IP, but not PA3.N 5 54B6321A7890 6 6785A3012B49 * 0123456 7 76941A3250B8 0 0123456 8 89AB01234567 1 1234560 9 987AB4503216 2 2356104 A AB4927856301 3 3410625 B BA5890741632 4 4561032 5 5602341 Figure 3.13. 12-element loop. PA, LALT, but not LIP, RIP, 6 6045213 RALT, FLEX, Antiaut.N

Feb 2004 4 3. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions Construction 3.14. LIP, LALT, but not PA, RIP, RALT, Figure 3.19. 12-element loop. PA, LIP, RALT, but not RIP, FLEX, Antiaut: Get a (6 · 27 = 162)-element example by LALT, FLEX, Antiaut.N taking direct product of 3.15 with 3.49. * 01234567 * 012345 0 01234567 0 012345 1 12345076 1 120453 2 23670145 2 201534 3 36701254 3 345012 4 45016723 4 453201 5 50167432 5 534120 6 67452301 7 74523610 Figure 3.15. 6-element loop. PA, LIP, LALT, but not LBOL, RIP, RALT, FLEX, Antiaut.N Figure 3.20. 8-element loop. RIP, RALT, but not PA, LIP, LALT, FLEX, Antiaut.N * 01234567 0 01234567 * 012345 1 12305674 0 012345 2 23016745 1 120534 3 30127456 2 201453 4 45670123 3 354012 5 54763210 4 435201 6 67452301 5 543120 7 76541032 Figure 3.21. 6-element loop. PA, RIP, RALT, but not Figure 3.16. 8-element loop. LBOL, but not RIP, RALT, RBOL, LIP, LALT, FLEX, Antiaut.N FLEX, Antiaut.N * 01234567 Construction 3.17. RIP, LALT, but not PA, LIP, RALT, 0 01234567 FLEX, Antiaut: Get a (12 · 27 = 324)-element example by 1 12307654 taking direct product of 3.18 with 3.49. 2 23016745 3 30125476 * 0123456789AB 4 45672301 0 0123456789AB 5 56741032 1 12305674AB98 6 67450123 2 2301674598BA 7 74563210 3 3012B8A97546 4 45672301BA89 Figure 3.22. 8-element loop. RBOL, but not LIP, LALT, 5 59B48A123607 FLEX, Antiaut.N 6 6B4A09285713 7 74A91B806325 * 0123456789ABCDEFGHJKL 8 8A9B74562031 0 0123456789ABCDEFGHJKL 9 9785A1B30264 1 123456789ABCDEFGHJKL0 A A678309B4152 2 23456789ABCDEFGHJKL01 B B856923A1470 3 3456E8FHB0D7JGACKL912 4 456789ABCDEFGHJKL0123 Figure 3.18. 12-element loop. PA, RIP, LALT, but not LIP, 5 56E89ABKD7FGHJCL01234 RALT, FLEX, Antiaut.N 6 6E8FABCL73GH9KDJ12045 7 7FGAJCDE12H4KL0893B56 * 0123456789AB 8 89ABCDEFGHJKL01234567 0 0123456789AB 9 9AB0DE3GHJKLF12645C78 1 12305674AB98 A ABCD7FG3JKL012H456E89 2 23019B8A6475 B BCDEFGHJKL0123456789A 3 30127456BA89 C CD7JGH95LF1234K0E86AB 4 4567012398BA D D7FGHJK6012345LE89ABC 5 56B812A97340 E E89HBKL0FG3J56712A4CD 6 6B4A83912057 F FGHCKLJ126450789AB3DE 7 78A639B25104 G GHJKL0123456789ABCDEF 8 8A9B67450213 H HJKL012A456E893BCD7FG 9 97852A0B4631 J JKL912045C786AB3DEFGH A A974B8301526 K KL01234C6E89AB5D7FGHJ B B459A0183762 L L012345DE89ABC67FGHJK

Feb 2004 5 3. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions Figure 3.23. 21-element loop. LRA, 2SI, but not PA, LIP, Figure 3.32. 6-element loop. PA, LIP, LALT, FLEX, xx = RIP, FLEX, Antiaut.N e, but not LBOL, RIP, RALT, Antiaut.N Construction 3.24. PA, LRA, but not LIP, RIP, FLEX, An- tiaut: Get a (14 · 12 = 168)-element example by taking direct Construction 3.33. RIP, RALT, FLEX, but not PA, LIP, product of 3.36 with 3.44. LALT, Antiaut: Get a (6 · 27 = 162)-element example by taking direct product of 3.34 with 3.49. * 012345 0 012345 * 012345 1 123450 2 230514 0 012345 3 345201 1 103452 4 451032 2 240531 5 504123 3 351024 4 425103 Figure 3.25. 6-element loop. FLEX, but not PA, LIP, RIP, 5 534210 LALT, RALT, Antiaut.N * 01234 Figure 3.34. 6-element loop. PA, RIP, RALT, FLEX, 0 01234 xx = e, but not RBOL, LIP, LALT, Antiaut.N 1 10342 2 24013 * 0123456789ABCDEFGHJKL 3 32401 0 0123456789ABCDEFGHJKL 4 43120 1 123456789ABCDEFGHJKL0 Figure 3.26. 5-element loop. PA, FLEX, xx = e, but not 2 23456789ABCDEFGHJKL01 LIP, RIP, LALT, RALT, Antiaut.N 3 345678FAB0DEJGHCKL912 Construction 3.27. LIP, FLEX, but not PA, RIP, LALT, 4 456789ABCDEFGHJKL0123 RALT, Antiaut: Get a (12 · 10 = 120)-element example by 5 56789ABCDEFGHJKL01234 taking direct product of 3.28 with 3.47. 6 678FABCDE3GH9KLJ12045 7 789ABCDEFGHJKL0123456 * 0123456789AB 8 89ABCDEFGHJKL01234567 0 0123456789AB 9 9AB0DE3GHJKLF12645C78 1 123074BA5698 A ABCDEFGHJKL0123456789 2 230189AB4567 B BCDEFGHJKL0123456789A 3 30125894BA76 C CDEJGH9KLF123450786AB 4 45670123AB89 D DEFGHJKL0123456789ABC 5 5894307612BA E EFGHJKL0123456789ABCD 6 698BA7052143 F FGHCKLJ126450789AB3DE 7 74BA16509832 G GHJKL0123456789ABCDEF 8 8BA563490721 H HJKL0123456789ABCDEFG 9 9A56B2387014 J JKL912045C786AB3DEFGH A A7492B816305 K KL0123456789ABCDEFGHJ B B6789A123450 L L0123456789ABCDEFGHJK Figure 3.28. 12-element loop. PA, LIP, FLEX, but not RIP, LALT, RALT, Antiaut.N Figure 3.35. 21-element loop. ALT but not PA, LIP, RIP, Antiaut. N Construction 3.29. LALT, FLEX, but not PA, LIP, RIP, RALT, Antiaut: Get a (6 · 21 = 162)-element example by taking direct product of 3.32 with 3.35. * 0123456789ABCD Construction 3.30. PA, LALT, FLEX, but not LIP, RIP, 0 0123456789ABCD RALT, Antiaut: Get a (6·14 = 84)-element example by taking 1 1234560A78DC9B direct product of 3.32 with 3.36. 2 2345601DA7B98C Construction 3.31. LIP, LALT, FLEX, but not PA, RIP, 3 345601289C7DBA RALT, Antiaut: Get a (6 · 27 = 162)-element example by 4 4560123CBD97A8 taking direct product of 3.32 with 3.49. 5 5601234BDAC879 6 60123459CB8AD7 * 012345 7 789ABCD0123456 0 012345 8 89C7DBA3016524 1 103254 9 9CB8AD76304215 2 240513 A A78DC9B1250643 3 354021 B BDAC8795462031 4 435102 C CBD97A84635102 5 521430 D DA7B98C2541360

Feb 2004 6 3. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions Figure 3.36. 14-element loop. PA, ALT, but not LIP, RIP, * 0123456789AB Antiaut.N 0 0123456789AB 1 12305674AB98 * 01234567 2 2301674598BA 0 01234567 3 30129B8A4657 1 12345076 4 45672301BA89 2 23017645 5 569B38A20714 3 35461702 6 69480A2B1375 4 46720153 7 7B86A9302541 5 50673214 8 8A74B0195263 6 67502431 9 945A12B87036 7 74156320 A A7B581936420 B B8A974563102 Figure 3.37. 8-element loop. Antiaut, PA3, but not PA, LIP, RIP, LALT, RALT, FLEX.N Figure 3.42. 12-element loop. PA, LRA, Antiaut, but not LIP, RIP, FLEX.N * 01234567 Construction 3.43. IPLR, but not PA, FLEX: Get a (12·18 = 0 01234567 216)-element example by taking direct product of 3.44 with 1 12056374 3.49. 2 20147635 3 34502716 4 47610253 * 0123456789AB 5 56371042 0 0123456789AB 6 63725401 1 123450BA7698 7 75463120 2 23450189AB67 3 3450127698BA Figure 3.38. 8-element loop. PA, Antiaut, but not LIP, RIP, 4 450123AB6789 LALT, RALT, FLEX.N 5 50123498BA76 6 6789AB234501 * 0123456789 7 789AB6501234 8 89AB67450123 0 0123456789 9 9AB678123450 1 1234567890 A AB6789012345 2 2356784901 B B6789A345012 3 3645978012 4 4768092135 5 5487109326 Figure 3.44. 12-element loop. PA, IPLR, but not LBOL, N 6 6579810243 RBOL, FLEX. 7 7890231564 8 8901623457 * 012345 9 9012345678 0 012345 1 123054 Figure 3.39. 10-element loop. IP, PA3, but not PA, LALT, 2 234501 RALT, FLEX.N 3 305412 4 450123 * 01234567 5 541230 0 01234567 1 12063754 Figure 3.45. 6-element loop. FLEX, Antiaut, but not PA, 2 20147635 LIP, RIP, LALT, RALT.N 3 34752016 4 45376201 5 56401372 * 012345 6 67520143 0 012345 7 73615420 1 103452 2 230514 Figure 3.40. 8-element loop. PA, IP, but not LALT, RALT, 3 345021 FLEX.N 4 451203 5 524130 Construction 3.41. LRA, Antiaut, but not PA, LIP, RIP, FLEX: Get a (12·21 = 252)-element example by taking direct Figure 3.46. 6-element loop. PA, FLEX, Antiaut, xx = e, product of 3.34 with 3.35. but not LIP, RIP, LALT, RALT.N

Feb 2004 7 3. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions * 0123456789 * 0123456789ABCDEFGH 0 0123456789 0 0123456789ABCDEFGH 1 1234567890 1 123450FHECGD79A86B 2 2356784901 2 2345018BA769HCGEFD 3 3465879012 3 345012EDHGFCB76A98 4 4578290163 4 450123A96B87DHFGEC 5 5687901324 5 501234GCFDEH9B86A7 6 6749018235 6 6D8CAH70B492FG531E 7 7890132546 7 7GBF9E062A4831HCD5 8 8901623457 8 8CAG6DB29074EF15H3 9 9012345678 9 9E7HBF4A082615CD3G A AH6D8C9472B0GE315F B BF9E7G28460A53DHC1 Figure 3.47. 10-element loop. IP, FLEX, but not PA, LALT, C CAH6D85G1E3F2079B4 RALT.N D D8CAH63E5F1G049B72 E E7GBF9D3C1H5A82046 F F9E7GBH1D5C386042A * 0123456 G GBF8E7C53HD16A4209 0 0123456 H H6D9CA1FG35E42B780 1 1205364 2 2014635 Figure 3.50. 18-element loop. PA, IPALT, but not LBOL, 3 3456210 RBOL, DIA.N 4 4631502 5 5362041 * 0123456789 6 6540123 0 0123456789 1 1032798465 2 2301987654 Figure 3.48. 7-element loop. PA, IP, FLEX, but not LALT, 3 3210879546 N RALT. 4 4798065132 5 5987604321 6 6879540213 * 0123456789ABCDEFGHJKLMNPQRS 7 7465132098 0 0123456789ABCDEFGHJKLMNPQRS 1 123456789ABCDEFGHJKLMNPQRS0 8 8654321907 2 23456789ABCDEFGHJKLMNPQRS01 9 9546213870 3 345678 J¨ AB MDEFGH¨ 9KL¨ CNPQRS012¨ 4 456789ABCDEFGHJKLMNPQRS0123 5 56789ABCDEFGHJKLMNPQRS01234 Figure 3.51. Unique 10-element Steiner loop. DIA, Commu- 6 678 JABCDE¨ Q¨ GH 9KLMNP¨ FRS012345¨ 7 789ABCDEFGHJKLMNPQRS0123456 tative, xx = e, but not LBOL, RBOL. The 12 Steiner triples 8 89ABCDEFGHJKLMNPQRS01234567 123 ¨ ¨ ¨ ¨ ¨ ¨ are the rows, columns, and generalized diagonals of 456 . N 9 9AB MD E QGHJKL 3NP 6RS012 C4 5 F7 8  789  A ABCDEFGHJKLMNPQRS0123456789 B BCDEFGHJKLMNPQRS0123456789A C CDEFGH 9¨ KL 3NPQRS012¨ M45678¨ JA¨ B D DEFGHJKLMNPQRS0123456789ABC * 0123456789AB E EFGHJKLMNPQRS0123456789ABCD 0 0123456789AB F FGH 9KLMNP¨ 6RS012345¨ Q78¨ JABCDE¨ G GHJKLMNPQRS0123456789ABCDEF 1 120534867B9A H HJKLMNPQRS0123456789ABCDEFG 2 201453786AB9 J JKL CN¨ P FRS012¨ M45¨ Q789AB¨ 3D¨ E 6G¨ H K KLMNPQRS0123456789ABCDEFGHJ 3 3450129BA687 L LMNPQRS0123456789ABCDEFGHJK 4 453201BA9876 M MNPQRS012 C45678¨ JAB¨ 3DEFGH¨ 9K¨ L N NPQRS0123456789ABCDEFGHJKLM 5 534120A9B768 P PQRS0123456789ABCDEFGHJKLMN 6 6789BA012354 Q QRS012345 F¨ 78 JABCDE¨ 6GH¨ 9KLMNP¨ R RS0123456789ABCDEFGHJKLMNPQ 7 786BA9201543 S S0123456789ABCDEFGHJKLMNPQR 8 867A9B120435 9 9AB687354012 A AB9876543201 Figure 3.49. The unique (< 36)-element IPALT but not PA B B9A768435120 loop. (Not PA since 1+8=96=J=3+6). Entries a ∗ b not agree- ing with integer addition a + b mod 27 have been decorated Figure 3.52. Unique 12-element non-associative Moufang ¨ with umlauts (M versus M). Note that these exceptions occur loop. N only on the index-3 subgrid and that the diagonal entries a+a, the first row and column 0 + a = a + 0, and the antidiagonal * 0 1 (−a)+ a = 0 never are umlauted. This loop has 27 elements Figure 3.53. 0 0 1 The 2-element group. N and is commutative, Lalt, Ralt, Flexible, LIP, RIP, antiaut, 1 1 0 but not power-associative, L-Bol, nor R-Bol. N

Feb 2004 8 4. 0. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions m m m m m m 4 Do LRalt loops have 2-sided in- G3 = sG1 = G1 s, G1 = sG3 = G3 s (6) verses? for all odd m ≥ 1. The question of whether LRalt loops have 2-sided inverses The effects of multiplying Ga powers by each other are (where sounds innocent. But it ushers us into a hurricane of com- n,m,j,k ≥ 0 always denote integers) plexity. Gj Gk = Gj+k, (7) In §4.1 we shall see that there are countably-infinite LRalt a a a loops without 2-sided inverses. However, there are no con- Gj Gk = GkGj = Gj+ksk, (8) tinuum-infinite analytically-smooth ones, since Sabanin [19] 0 2 2 0 0 j k k j j+k k showed that analytic LRalt loops are diassociative. G1G3 = G3 G1 = G1 s , (9) In §4.2 we examine the evidence suggesting that inverses are − Gn m if m ≤ n always 2-sided in any finite LRalt loop. m n  1 G0 G1 =  m−n (10) In §4.5 we suggest that the LRalt=⇒2SI problem may actu- G0 if m ≥ n ally be among the hardest mathematical problems that are  n−m m this simply posed. G0 s if m ≤ n GmGn =  (11) 1 0  m−n n G1 s if m ≥ n 4.1 A countably infinite LRalt loop without  2-sided inverses It is now a straightforward matter to see that both left- and right-division are uniquely defined, so that we indeed have a All the elements of the loop may be defined in terms of 6 loop, and that Lalt and Ralt indeed are obeyed. particular elements we call e, s, and G0, G1, G2, G3. Power-associativity is obeyed. The antiautomorphic inverse The loop will obey Lalt xx·y = x·xy, and Ralt x·yy = xy ·y. property (x\1)(y\1)=(yx)\1 is false in this loop. Indeed we The identity element is e so that xe = ex = x for all x. The do not have any antiautomorphism, nor does the loop obey special element s (which also may stand for either sign or LIP nor RIP, since any of these would have caused 2-sided swap) obeys inverses. • xs = sx (s commutes with everything); However, the following three maps all are automorphisms: • s · s = e (s is self-inverse; consequently sk = s or e if k x → 1/x (which maps G → G − ), x → x\1 (which is odd or even respectively); a a 1mod4 maps G → G ) and x → 1/(1/x) (or x → (x\1)\1, • Hence as a consequence of Lalt, ss·x = s·sx = s·xs = x, a a+1mod4 which in this loop happens to be the same map; note that this and as a consequence of Ralt, x · ss = xs · s = sx · s = x map is involutive) which swaps G ↔ G . (thus multiplication by s is a self-inverse operation); a a+2mod4 • if xy = z then sx · sy = z and sx · y = x · sy = sz, Osborn’s [15] weak inverse property y((xy)\1) = x\1 is (multiplication by s has interesting “pairing”effect; also obeyed in this loop; thus both WIP and the automorphic in- s associates with everything); verse property hold, which is often called the crossed inverse • if xy = 1 and zx = 1 then either y = z or {ys = sy = z property CIP. and zs = sz = y} (if x has two unequal one-sided in- An A-loop is a loop whose inner mappings (i.e the identity- verses, then s-multiplication interchanges them); preserving permutations of the loop’s elements induced by • either xy = yx or xy = yxs (near-commutativity). compositions of left- and/or right-multiplications) all are au- tomorphisms. Our infinite loop is not an A-loop because its G0, G1, G2, G3 obey inner mapping x → xG0 · G1 is not an isomorphism. G0 = sG2 = G2s, G2 = sG0 = G0s, (1)

G1 = sG3 = G3s, G3 = sG1 = G1s (2) 4.2 Do finite LRalt loops have 2-sided in- and verses? 7 GaGa+1 mod 4 = e (3) Exhaustive searches with mace4 show that every LRalt loop so that each of them has two distinct 1-sided inverses. (indeed, every LRalt magma with \-division and x1= x) with ≤ 185 elements has 2-sided inverses. The full set of elements of the loop are Define n m n m 2n 2n n def {G0 , G1 , G2 , G3 , sG0 , sG1 , e, s}, n,m ≥ 1, m odd. Xℓ = X(X(X(X...X))), (12) (4) n X’s in all The reason we said that m had to be odd was to prevent | {z } 2k 2k 2k 2k i.e. Xn denotes the result of starting with 1 and doing a left- element-duplication, because G2 = G0 and G1 = G3 if ℓ m m m m multiplication by X repeatedly n times. (It was this leftward k ≥ 0. (Similarly, G0 s = G2 and G1 s = G3 if m is odd.) kind of exponentiation that was intended in the abstract.) We The remaining effects of multiplying these elements by s are shall later also have use of Xn, which is defined similarly but covered by the facts that s associates and commutes with ev- r using right-multiplication; and we shall use Xn without any erything and that subscript when we intentionally wish to leave its parenthe- m m m m m m G2 = sG0 = G0 s, G0 = sG2 = G2 s, (5) sization ambiguous. 7It is necessary to modify the source code to permit loops with over 100 elements. Mace4 reached 185 in only 1 day and then stopped because it ran out of memory.

Feb 2004 9 4. 2. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions Lemma 4 (Exponents of finite loops exist). Let X be xy · (y · y(yy)) = (x · y(y · yy))y (16) an element of a finite loop L. Then there exists a positive in- n then letting y = x in the latter and applying Lalt to its left teger n, called its “left-exponent,” such that Xℓ =1. Further, there exists N (the “left-exponent of the loop”) such that for hand side. (All 4 of these identities arise solely from the LRalt N axioms.)  all U ∈ L, Uℓ =1. Proof: The repeated left-multiplication process must by One might now conjecture that “the pattern continues” in the n−1 finiteness ultimately repeat a value. Suppose the first repeat sense that we somehow may always convert X · Xℓ into n− a b a b X 1X by “playing with parentheses,”i.e. by using the LRalt is Xℓ = Xℓ with 0 ≤ a

6 The n = 6 case may be proven by showing the identities Figure 4.2. 21-element LRalt quasigroup in which 0 ∗ 0ℓ = 6 0 ∗ 5=6 6= D = 5 ∗ 0=0ℓ ∗ 0. This is in fact a loop with (x · yy)y = (xy · y)y = xy · yy, (15) identity element L. N R R 8That is, among the usual loop axioms alone, we cannot omit the demand that at least one kind of division exists, and we cannot omit the demand that the quantity 1 such that x−1x = 1 defines left-inverses, must in fact be a left-identity. However, we might, conceivably, be able to replace these “non-omittable” axioms with some other, less-usual, statement. Feb 2004 10 4. 2. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions * 01234567 Lemma 9. If n = 2k +2j , then n is mirrorable in LRalt 0 01234567 magmas. 1 12345671 n− Proof: When n =2k + 1 lemma 7 shows that Xn = X 1X. 2 23456712 ℓ ℓ But since n − 1 is a power of 2, lemma 9 shows this is 3 34560123 n−1 n = Xr X = Xr . 4 45671234 k n n−2 5 56012345 When n = 2 + 2 lemma 7 shows that Xℓ = Xℓ · XX. 6 60123456 But since n − 2 is a power of 2, lemma 9 shows this is n−2 n−2 n = Xr · XX = Xr X · X = Xr . 7 71234560 R We may indeed use lemma 9 to double 2k−j + 1 repeatedly j Figure 4.3. 8-element magma with identity e = 0 but with- k j 6 times to get that 2 +2 is mirrorable for any j with 0 ≤ j ≤ k. out left- or right-division. Obeys LRalt but 3 ∗ 3ℓ = 3 ∗ 4 =  6 N 0 6=7=4 ∗ 3=3ℓ ∗ 3 so 2SI is false.

k k Lemma 10. Let n> 1 be odd. A necessary condition that ei- Nevertheless, the conjecture is true if n =2 +1 or n =2 +2: n−1 n−1 ther X ·Xℓ = Xℓ X or that n be mirrorable, in any LRalt n k magma, is that n divide some 2k +2j with 0 ≤ j ≤ k. For the Lemma 7 (LRalt=⇒2SI if Xℓ =1 where n − 2 =1, 2). If n =2k +1 or n =2k +2 then each element X in an LRalt former problem, it is necessary in addition that 0 ≤ j < k if n−1 n−1 k ≥ 2. These conditions in general remain necessary even if magma obeys XXℓ = Xℓ X. the LRalt magma is known to have an identity element e and Proof: Consider the expression n it is known that Xℓ = e. X(X[X(X[X · · · Xy])]) (17) Proof: Since n> 1 is odd, for any a,b > 0 with a + b = n we have, without loss of generality, 0

k k  (2 ) (2 ) counterexample. where Xc is X parenthesized in the manner of a complete depth-k binary tree. The above lemmas show that the following n with 1 ≤ n ≤ 20 We now use the equality of EQs 17 and 21 in the cases y = X are mirrorable: 1,2,3,4,5,6,8,9,10,12,16,18,20, while the fol- − − and y = 1. The result is that XXn 1 = Xn 1X if n =2k +1. lowing n are not mirrorable in general LRalt magmas: 7,15. ℓ ℓ Further, 7 is not even mirrorable in LRalt loops due to the To now consider n = 2k + 2, let y = XX and find that EQ k 21-element counterexample in figure 4.2. Mace4 also found ex- n (2 ) 17 is just Xℓ while EQ 21 is Xc · XX. This by Ralt is plicit LRalt magmas with 1 in which 11,13, are not mirrorable. k (2 ) k Xc X · X and by the preceding result about 2 + 1 this is Nevertheless, I conjecture that 11 and 13 are mirrorable in n−1  magmas with right-division (exhaustive searches show that just Xℓ X. n n any counterexample must have > 185 elements) and indeed “Mirrorable” n: Define n to be mirrorable if Xℓ = Xr in that: an LRalt loop, or magma, or whatever we are talking about at the moment of use. Conjecture 11 (Mirrorability). An integer n> 0 is mir- rorable in an LRalt magma with right-division if n divides Lemma 8. If n is mirrorable in LRalt magmas, then so is some number of the form 2k +2j where 0 ≤ j ≤ k. 2n. 2n n Lemma 12. Let n be the least common multiple of the left- Proof: By making an Lalt pass, Xℓ = (XX)ℓ By assump- n 2n  and right-exponents of an element X in an LRalt loop, i.e. let tion this is (XX)r which by an Ralt pass is Xr . n n n > 0 be the least integer such that Xℓ = Xr = 1. (If the Consequently, by induction, powers of 2 are mirrorable. A loop is finite, such an n always exists.) Let g, j, k ≥ 0 and let j k m m more general statement is m =2 [2 + 1] − gn > 0. Then Xℓ = Xr . Feb 2004 11 4. 2. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions Proof: 2j[2k + 1] is mirrorable by previous results, and then Conjecture 14. If n divides some number of the form we may simply “chop off” g chunks of n X’s from the prod- 2k + 2j where 0 ≤ j < k, then each element X in an n n n−1 ucts Xℓ and Xr on the grounds that multiplying by 1 has no LRalt magma with left-identity obeying XXℓ = e neces-  n−1 effect. sarily obeys Xℓ X = e.

Lemma 7 suffices to get quite far. (Incidentally, it is not hard to prove that the sets of numbers Lemma 13. If n> 0 is any integer which divides some num- obeying the conditions in conjectures 11 and 14, while infinite, k n−1 n−1 contain arbitrarily large “gaps.”) ber of the form 2 +2, then 1= X ·Xℓ implies 1= Xℓ ·X in an LRalt magma with identity. Finally, neither otter nor I were able to handle n = 14, 15, In particular, this criterion includes directly, although 14 eventually succumbed to an indirect at- tack. Specifically, we shall provide proofs for 14 and 15 under 1. All primes not congruent to 7 mod 8 (but primes con- the assumption of conjecture 11. Later otter was able to set- gruent to 7 mod 8 are excluded), tle that conjecture in 13-case, providing a proof for n = 14. 2. All n which factor into primes congruent to 3 mod 8 (for example n =99=3 · 3 · 11), Unfortunately, otter’s proofs get more and more complicated 3. Among the n with 2 ≤ n ≤ 20, precisely the following: with increasing n and lack any recognizable pattern. 2, 3, 5, 6, 9, 10, 11, 13, 17, 18, 19. Otter produced a spectacularly complicated 88-step proof for n kn Proof: Because Xℓ = 1 implies that Xℓ = 1 we have that the case n = 7. Otter is a computerized deduction engine by kn−1 n−1 kn−1 Xℓ = Xℓ so that it suffices to prove Xℓ X = 1. In W.McCune. One may input axioms to it (e.g. the LRalt other words, “if it works for some multiple kn of n, then it and loop axioms) and ask to to prove some desired conclusion works for n.” We now sketch the proofs of the specific cases: (e.g. that 1/x = x\1). In some cases, otter will succeed 1. We have already dealt with p = 2. So let p be an in finding a proof; in others it will run out of time or mem- odd prime. Then it follows from Gauss’s quadratic reci- ory. Sometimes otter can be far inferior to a human math- procity theorem that some power of 2 is congruent to ematician. Other times – favorable circumstances are when −1 mod p, i.e. p divides 2k +1 for some k, if and only there are few axioms and little human-exploitable “structure” if p is not congruent to 7 mod 8. – otter seems to achieve vastly superhuman deductive power. 2. If n factorizes into primes congruent to 3 mod 8, then This is one of them: otter found its proof for the case n =7 some power of 2 is congruent to −1 mod n because 2k in 17 seconds, and similar proofs for all cases we’ve mentioned will do where k is the least common multiple of the in- combined in under 10 minutes. I do not believe any human dividual k’s; note this will always be an odd multiple of can match that performance9. Indeed, this human was un- each. able even to fully understand otter’s n = 7 proof. Even 3. 26 +2=66=2 · 3 · 11, 27 +2 = 130 = 2 · 5 · 13, single deductive steps in an otter proof can be quite non- 210 +2=2 · 9 · 19, 213 +2=2 · 17 · 241. But 7 and hence trivial, e.g. “paramodulations” with many parameters. For 14 are excluded by claim#1; 8, 12, 16, and 20 obviously example, according to otter’s notion of a “single step,” set- cannot divide any 2k + 2; finally 15 does not divide any tling the case n = 6 (as we did above) requires only 2 steps! 2k +1 because 3|(2k + 1) only when k is odd, whereas Thus really, otter’s “88-step” proof perhaps would be more 5|(2k + 1) only when k ≡ 2 mod 4. properly regarded as a 200-300 step proof.

 Nevertheless, the honor of humanity ultimately partially re- asserted itself when I produced the following much simpler The criterion of lemma 13 admits a fairly large set of integers n = 7 proof. It comes quite easily once one adopts the goal of n. The number of primes not congruent to 7 mod 8 below incorporating both the identity element 1, and cancellation, x is asymptotic to 0.75x/ ln x. The set of n with 1

Feb 2004 12 4. 2. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions First otter derived the following 5 identities from the LRalt have 2-sided inverses for both n = 14 and 15, by: writing a axioms alone: special purpose computer program based on standard graph- connectivity algorithms and the graph-reformulation of the [x · y(y · yy)]y = xy · [y(y · yy)] (22) problem in §4.3. If so, then the next open case would be n = 21. [(x · xy)(x[x · yy])]y = (x · xy)[(x · xy) · yy] (23)

[xy · (x · yy)]y = xy · [xy · yy] (24) 4.3 The graph picture

[x · x(x · xy)]y = x · x[x(x · yy)] (25) Let Gn be the graph whose vertices consist of the ordered 3 5 8 rooted binary trees with n leaves. Each such tree represents a xℓ xℓ = xℓ . (26) n way to parenthesize X . The edges of Gn join two trees which Various facts are true about A which are not true for gen- are equivalent by an Lalt or Ralt re-parenthesization. The eral x. For example, A · 1 = A, even though we had only question of whether an element X in an LRalt magma obeys n−1 n assumed 1 was a left identity. This arises from the identity Xℓ X = Xℓ , is then equivalent to the question of whether 8 x · x(x · x[x · x(x · xy)]) = xℓ y (which is a special case of the these two particular vertices of Gn lie in the same connected 9 8 proof of lemma 7) by using y = x = A to get Aℓ = Aℓ A and component. This view enables proving certain statements eas- then recognizing the left hand side as A · 1 and the right as ily, which otherwise might have been difficult. For example 1 · A = A. Lemma 15 (Unboundedly large proof lengths). The Note that we derived A1= A by starting with some generally graphical distance (number of edges in the shortest path be- true identity applied to A, left-multiplying various subexpres- tween) these two vertices is at least n − 2. sions by some form of 1 = A8, rearranging parentheses, and then recognizing certain other subexpressions as forms of 1 Proof: Each n-leafed ordered roted binary tree may be re- and hence removing them. Otter used this same strategy (but garded as the planar dual of a triangulation by diagonals of 3 3 6 4 3 7 in more complicated ways) to find Aℓ Aℓ = Aℓ , Aℓ Aℓ = Aℓ , a convex (n + 1)-gon with one distinguished “root edge” AB. 6 6 4 5 6 3 3 6 3 4 3 6 Aℓ Aℓ = Aℓ , Aℓ Aℓ = Aℓ , Aℓ Aℓ · Aℓ Aℓ = 1, Aℓ Aℓ = A, Each associative transformation corresponds to erasing a di- 3 4 7 7 8 Aℓ Aℓ = Aℓ , and the goal of the proof, namely Aℓ A = Aℓ =1 agonal and then retriangulating the resulting quadrilateral- (as well as many other, less simply expressible, claims). All of shaped hole using the other diagonal (and LRalt transforma- these are true for A but are unobtainable (in LRalt magmas tions are a subset of associative transformations). Originally with left-identity) for general x. all the n − 2 diagonals have endpoint A but none have end- The finale of otter’s proof is as follows. It manages to ob- point B, but in the final state, the opposite is true, so at least 3 6 3 4 n − 2 transformations have to be made.  tain Aℓ Aℓ · Aℓ Aℓ = 1. It then uses this fact (among others) to derive A3A4 = A7. From this we know (A3A4)A = A7A. ′ ℓ ℓ ℓ ℓ ℓ ℓ Now define the graph Gn, which has an infinite number of Now applying EQ 23 with x = y = A to the left hand side vertices, as follows. Its vertices correspond to the ordered 8 7 gives Aℓ = Aℓ A and upon recognizing the left hand side as 1 rooted binary trees with kn leaves for all k = 1, 2, 3,... . In  we have proven the theorem. addition to the LRalt edges we mentioned before, there are 14 13 also edges linking kn-leaf trees to (k + 1)n-leaf trees, corre- n=14: To prove Xℓ = 1 implies Xℓ X = 1 in an LRalt loop in which 13 is mirrorable: Left-multiply by XX and sponding to multiplying some subexpression on the left or n 16 right by Xell = 1. The question of whether an element X employ Lalt to get Xℓ = 1. Now by repeated uses of Lalt to pair X’s into XX’s, then into X4’s, and so on we have in an LRalt magma with 2-sided identity 1 has a 2-sided in- c n 16 8 4 4 8 2 16 verse, given that Xℓ = 1, is then equivalent to the question Xℓ = (XX)ℓ = (Xc )ℓ = (Xc ) = Xc so that now by a ′ 16 16 16 14 14 of whether two particular vertices of G lie in the same con- mirror argument Xℓ = Xc = Xr = Xr X · X = Xr · XX. n R 14 14 nected component. Now since Xℓ = 1 we have that XX = Xr · XX so that by ′′ 14 13 Finally, define the graph Gn. It too has an infinite num- cancelling the XX we get Xr = 1. This is Xr X = 1. Now 13 13  ber of vertices. Now they correspond to the unordered pairs if 13 is mirrorable, then Xr = X and we are done. ℓ of ordered rooted binary trees with finite numbers of leaves. Remark. M.K.Kinyon (private communication) was able to In addition to the LRalt and 1-multiplication edges we men- get otter to prove 11 and 13 mirrorable in loops. That com- tioned before (now operating independently on each member ′′ ′ ′ pletes the n = 14 proof above. of the pair; so far Gn = Gn × Gn but we shall now adjoin an 15 14 n=15: To prove Xℓ = 1 implies Xℓ X = 1 in an LRalt loop infinite number of additional edges), there are also edges link- 29 in which 29 is mirrorable: It suffices to prove Xℓ X = 1. ing pairs of trees to pairs of trees each with K extra leaves, 32 Left-multiply by XX and employ Lalt to get Xℓ = 1. Now corresponding to multiplying both entire expressions by some by repeated used of Lalt to pair X’s into XX’s, then into common K-term expression on the left or right, i.e. (in the 4 Xc ’s, and so on we get (similarly to in the previous proof) latter case) to adjoining new roots to both trees in the pair, 32 32 32 30 30 Xℓ = Xc = Xr = Xr X · X = Xr · XX. Now since whose two left children are the two old trees, and whose two R 30 30 right children are two copies of the same new K-leaf tree. Xℓ = 1 we have that XX = Xr · XX so that by cancelling 30 29 The question of whether an element X in an LRloop has a the XX we get Xr = 1. This is Xr X = 1. If 29 is mir- n 29 29  2-sided inverse, given that Xℓ = 1, is then equivalent to the rorable then Xr = Xℓ and we are done. ′′ question of whether the two particular vertices of Gn repre- n n−1 It probably would be possible to establish fully rigorously the senting {Xℓ , 1} and {Xℓ X, 1} lie in the same connected fact that elements X with left-exponent n in an LRalt loop component.

Feb 2004 13 4. 4. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions Connectivity questions about infinite graphs or directed For the purpose of comparison, consider these 8 problems: graphs are notoriously difficult, the “3x+1 problem” [8] being a simple prototypical unsolved example. 1. What is the maximum number of faces of a convex poly- hedral space-tiler? 4.4 Possible 87% solution? 2. Does white always win a perfectly played chess game? 3. Are there an infinite number of primes of the form The situation so far is: The LRalt=⇒2SI problem remains n2 + 1? unsolved in finite loops. Even for the (apparently simpler) 4. The“3x+1 problem” [8] of whether the iteration on the problem in magmas with left-identity, or magmas with right- positive integers x → 3x + 1 if x is odd, x → x/2 if x is division, we have been unable to completely settle the ques- even, will always reach the value x = 1, no matter what n n n tions of which n cause Xℓ = Xr and which n cause Xℓ =1 n−1 the starting point. to imply Xℓ X = 1. But we have made some progress by 5. Does P=NP? [6] finding some necessary, and some sufficient conditions. 6. Are all planar graphs 4-colorable? [17] We now sketch a plan of argument which may enable an 7. Classify all finite simple groups. [3] “87.5% solution.” A conjecture essentially the same as the 8. Do there exist integers a,b,c > 0 and n > 2 with standard conjecture that there are an infinite number of twin an + bn = cn? [22] primes (p,p +2) is Conjecture 16 (Modified twin-primes). Given an odd All are mathematically natural with the possible exception of number n, there are an infinite set of numbers k such that the 3x + 1 problem. 4kn +1 and 4kn − 1 both are prime. The first 5 problems (with the possible exception of the first, Theorem 17. Let n > 0 be an integer not divisible by 8. which has not been worked on as hard as the others) seem Under the assumption of conjectures 11 and 16: in an LRalt well beyond reach, whereas problems 6-8 have (with immense n magma with identity 1 and with right-division, X = 1 im- effort) been solved, albeit the solutions of 6 and 7 seem suffi- plies that X has a 2-sided inverse. ciently long that it is almost beyond the ability of any single Proof: Let n not be divisible by 8. Find k so kn ± 1 both are human to check them. prime, and kn ≡ 4 mod 8. Then observe that by conjecture 11 and some easy number theory, that kn±1 both are mirrorable. Although all of these problems may seem simple to state, kn kn+1 kn+1 one gets a more precise perspective on the “simplicity” of a Left-multiply Xℓ by X to get Xℓ = Xr = X. Right- kn kn−1 problem statement after gaining some experience inputting cancel the X’s to get Xr =1= Xr X. Now mirror to − − get = Xkn 1X =1= Xn 1X. problems to computerized proof- and counterexample-finding ℓ ℓ tools such as otter and mace. If one is only allowed to em- This would totally settle the LRalt=⇒2SI problem except for ploy first order logic and must describe the problems to a those n that are multiples of 8 (i.e. 7/8=0.875 of all n). completely naive listener, i.e. such a computerized system, Further, some of the multiples of 8 could be handled by con- then the descriptions of each of these 8 problems are longer jecture 14; the first open case would be n = 56.  – usually much longer – than ours. (Just getting started by This proof-plan, of course, still would only provide “87% of a defining notions such as “integers,” “primes,” “space-tiling,” solution,” and cannot be implemented at least until the 3000- “convex polyhedron,” “planar graphs,” “classification of sim- year-open twin-primes problem is settled! In that case, theo- ple groups,”“P,”“NP,” or the rules of chess already takes too rem 17 merely would serve to reduce the problem to proving long.) Probably the simplest 3 to state among our 8 are the conjectures 14 and especially 11. Still, that reduction ar- 3x + 1 problem, the classification of simple groups, and Fer- guably is progress since these conjectures concern LRalt mag- mat’s last theorem, but neither is as simple as ours. mas with left-identity and right-division, respectively (i.e. not It is notoriously hard to estimate the difficulty of open prob- both at the same time) – more simply defined objects than lems, and especially so in the case of LRalt=⇒2SI since LRalt loops. M.K.Kinyon11 and I are the only people who have tried hard on it. But let us say this. 4.5 Candidate for the most frustrating prob- 1. The (n2 + 1)-primes and twin-primes problems have been lem in the world? open for thousands of years, which makes them harder than the last three (solved) problems. We’ve seen reasons to sus- Connoisseurs of frustration prefer their problems to be sim- pect the LRalt=⇒2SI problem may be at least equally hard ple to state, mathematically natural, and difficult to solve. as them and the 3x + 1 problem. According to these criteria, the LRalt=⇒2SI problem is a 2. Everybody is confident they already know the answers to surprising contender for the world’s top problem! problems 3,4,5 but merely cannot prove it; and it is trivial The LRalt=⇒2SI problem has an extremely simple statement: to investigate 3,4, and 8 up into the billions by computer. Given that, in a finite universe, In contrast, computer investigation of LRalt=⇒2SI is much harder and I feel considerably less confidence I know its an- 1 ∗ x = x; (x/y) ∗ y = x; (27) swer. 3. Solving chess is a “trivial”problem in that it may be solved x∗(y ∗y) = (x∗y)∗y; (y ∗y)∗x = y ∗(y ∗x); (28) purely mechanically by a finite case analysis using only first does it then follow that x ∗ (1/x)=1? order logic. But (we have shown in §4.1) no such solution is

11See §5. Feb 2004 14 4. 5. 0 Smith typeset 16:23 30 Jan 2005 loop inclusions possible for LRalt=⇒2SI (at least, if the answer, as expected, [3] J.H.Conway, R.T.Curtis, S.P.Norton, R.A.Parker, R.A.Wilson: F ATLAS of Finite Groups, now reprinted with corrections and ad- is positive). ditions Oxford Univ. Press, November 2003. Should the joys of LRalt=⇒2SI pall, the reader is reminded [4] Vilnis Detlovs & Karlis Podnieks: Introduction to Mathematical that there are 5 more problems of the same ilk listed in table Logic, electronically available textbook at University of Latvia, 1.1. All of them are only slightly harder to state and they http://www.ltn.lv/∼podnieks/mlog/ml.htm. Chapter 4 covers may be even harder to solve. completeness theorems. [5] Herbert B. Enderton: A mathematical introduction to logic, Aca- 5 Acknowledgements and updates demic Press 1972. [6] M.R. Garey & D.S. Johnson: Computers and Intractability, J.D.Phillips found a few of the loop examples in §3, or closely W.H.Freeman and Company, San Francisco, 1979. related loops, before I did, and also did some work on the [7] Kurt G¨odel: Die Vollst¨andigkeit der Axiome des logischen LRalt=⇒2SI problem, which I had dropped on him rather Funktionen-kalkuls,¨ Monatsh. fur¨ Mathematik und Physik 37 like a bomb. We both initially thought that problem was (1930) 349-360. going to be far easier than it now seems. It was he who ini- [8] Jeffrey C. Lagarias: The 3x+1 problem and its generalizations, tially suggested the conjecture that its solution depends on Amer. Math. Monthly 92 (1985) 3-23. the finiteness of the loop; all my investigations so far support that. [9] W.McCune: Son of Bird Brain (automated deduc- tion system demonstration web page), http://www- P.Vojtechovsky spotted a typo which caused the appearance unix.mcs.anl.gov/AR/sobb/. of a serious error. (It has been corrected.) [10] William W. McCune: Mace 2.0 reference manual and guide, It should be obvious that I have made heavy use of cs.SC/0106042 mace4 [10] and otter [11]. My own programs loop- beaut.c and setinc.c, are available on my website [11] William W. McCune: Otter 3.3 reference manual, cs.SC/0310056 http://math.temple.edu/∼wds/homepage/works.html. [12] W. McCune & R. Padmanabhan: Automated Deduction in Equa- The use of all 4 of these programs (albeit setinc.c would tional Logic and Cubic Curves, Springer-Verlag (LNCS [AI sub- have to be appropriately modified) should enable future in- series] #1095) 1996. vestigations of the same sort to proceed in a highly automated [13] W. McCune: Solution of the Robbins Problem, way. J.Automated Reasoning 19,3 (1997) 263-276 and http://www- Updates: M.K.Kinyon and I (with computer aid) have ex- unix.mcs.anl.gov/∼mccune/papers/robbins/. amined the LRalt=⇒2SI problem further since this paper was [14] Elliot Mendelson: Introduction to mathematical logic, Lewis Pub- written and conceivably a followup paper by one or both of us lishers Inc. 4th ed. 1997. may appear. In particular Kinyon showed that X must have [15] J.Marshall Osborn: Loops with the weak inverse property, Pacific n a 2-sided inverse in an LRalt loop if X = 1 with n ≤ 31 and J. Math. 10 (1960) 295-304. n = 63 and I showed this is true in an LRalt magma with k [16] J. D. Phillips and Petr Vojtechovsky C-loops: An Introduction, 2-sided identity if n divides any number of the form 2 + 2. M04/03 at http://www.math.du.edu/preprints.html. Kinyon suggests that a possibly-productive line of attack on the LRalt=⇒2SI problem would be to prove in an LRalt [17] N.Robertson, D.P.Sanders, P.D.Seymour, R.Thomas: The four colour theorem, J. Combin. Theory B 70 (1997) 2-44. magma with 1x = x1 = x, that An = 1 implies these two − − ℓ − − n 1 2 n 1 n 1 n 1 [18] D.A.Robinson: Bol loops, Trans.Amer.Math.Society 123 (1966) equations hold: Aℓ A = A, Aℓ (Aℓ A) = Aℓ . If the magma has right or left cancellation (respectively), then these 341-354. respectively would suffice to imply 2SI. Kinyon also suggests [19] L.V.Sabanin: On the diassociativity of smooth monoalternative − − − − − n 1 n 1 n 2 n 1 n 1 maps, Russian Math. Surveys 51 (1996) 747-749. investigating Aℓ Aℓ = Aℓ and (Aℓ )ℓ = A. [20] Warren D. Smith: Loop diassociativity has no finite basis, avail- References able at http://math.temple.edu/∼wds/homepage/works.html. [21] Warren D. Smith: Quaternions, octonions, and now, 16-ons and n [1] G.Bol: Gewebe und Gruppen, Math. Annalen 114 (1937) 414-437. 2 -ons; New kinds of numbers. Book, under review for publica- tion. [2] Richard Hubert Bruck: A survey of binary systems, Springer- Verlag 1958, third corrected printing 1971 (Ergebnisse der Math. [22] Andrew Wiles; Modular elliptic curves and Fermat’s Last Theo- #20). rem, Annals of Mathematics 141 (1995) 443-551.

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