Word for group of three

What is the word for a group of three?

In this page you can discover 44 synonyms, antonyms, idiomatic expressions, and related words for trio, like: set of three, triad, troika, vocal trio, trinity, triangle, triplet, triplicate, threesome, trey and null.

What is it called when a group of people?

Originally Answered: What is a group of humans called? for general humans, the collective noun is a ‘crowd’ of humans. List of Collective Nouns for People + Groups of Humans + Professions.

What do you mean by Trio?

1a : a musical composition for three voice parts or three instruments. b : the secondary or episodic division of a minuet or scherzo, a march, or of various dance forms. 2 : the performers of a musical or dance trio. 3 : a group or set of three.

What triumvirate means?

1 : a body of triumvirs. 2 : the office or government of triumvirs. 3 : a group or association of three.

What is group of four called?

A group of four, a quartet or a game (such as golf) played by four players, especially by two teams of two. ensemble. foursome. group. quartet.

What is a good name for a group?

To help get you started, here are some universally-liked team names that could be your group’s name.A Team.All Stars.Amigos.Avengers.Bannermen.Best of the Best.Bosses.Champions.

What is the best group name for friends?

Friends Group Chat NamesKylie Is Our Mother.The Meme Team.Best Fries Forever.The Friendship Ship.The Chamber of Secrets.F is For Friends Who Do Stuff Together.The Real Housewives of ______Taylor Swift’s Squad.

What is another name for 4?

Find another word for four. In this page you can discover 41 synonyms, antonyms, idiomatic expressions, and related words for four, like: 4, little-joe, quatern, quadrigeminal, quadripartite, foursome, quaternity, quartet, quadrisect, four-spot and quaternary.

What is a good group name for 5 friends?

65 Group Chat Names For 5 Best Friends, Because They’re Your Main BeachesResting Beach Faces.Mermaid To Be Five.Power Rangers.Spice Girls.Backstreet Girls.Always *NSYNC.Ride Or Dies.My Lucky Charms. AS photo studio/Shutterstock.

How do you name a team?

7 Great Ways to Come Up With a Memorable Sports Team NameDecide If Your Want Sport Type In The Name. Associate Your Team With Popular Things. Think About Things Your Team Members Have In Common. Add a Strong Adjective. Pair a Mascot With a Location. Use a Sports Team Name Generator. Make Sure All Team Members Are On Board.

Whats a good name for a duo?

Dog Duo Names For Girls or BoysAbra & Cadabra.Alpha & Beta.Bangers & Mash.Basil & Parsley.Button & Bow.Cheese & Cracker.Cookie & Cream.Fish & Chip.

What is the most iconic duo?

The Most Iconic Duos In Movie HistoryBrennan Huff and Dale Doback – Step Brothers. Hannibal Lector and Clarice Starling – Silence of the Lambs. Doc and Marty – Back to the Future. Butch Cassidy and the Sundance Kid. Jules and Vincent – Pulp Fiction. Harry and Lloyd – Dumb & Dumber. Kirk and Spock – Star Trek. Han Solo and Chewbacca – Star Wars.

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On reading the title of this post, readers would have immediately been able to provide the answer and may have wondered why I was even asking it.

But as I was writing my impressions of Lucia di Lammermoor and the sextet that is sung there, it struck me that while I knew the names for groups of singers of almost all sizes from two to ten (duet (two), quartet (four), quintet (five), sextet (six), septet (seven), octet (eight), nonet (nine), and dectet (ten)), I did not know the term for three singers. So I looked it up and (duh!) it is ‘trio’, a common word that I was very familiar with. After all, the Kingston Trio was a very popular group in my youth.

So why could I not recall this very obvious word? The answer may lie in the fact that the human brain retrieves information by following links from whatever starting point triggered the memory process. Since my curiosity had been piqued by the word ‘sextet’ and branched out from there, it probably began searching for a word that ended with ‘et’ in order to fit the pattern and it appears that there is none.

It is interesting how the brain makes associations, so that a piece of information that one cannot recall because it is being searched for in one context is easily recalled in another. If I had been thinking of other forms of music, such as folk or jazz, then the word ‘trio’ would have come to mind immediately since they are so commonly used in those genres.

I don’t know why no word ending with ‘et’ was coined for groups of three but then English is full of exceptions to patterns. Maybe there had been a word at some time (‘triet’?) that slowly disappeared from common use.

• #1

Hi Everybody.

I am wondering if, in the same of «tupla» indicating a group of two things, there is a word to say a group of three. In Spanish, we have «tripleta» but I don’t have a clue if there is such a word in English.

Thanks a lot for your help
Cristinaalp

• #2

Tupla? This is not an English word, it is Finnish (meaning double). Do you mean ‘couple’ for a group of two?
We have triplet in English, it usually refers to three children who are born of the same mother, at the same time. However it can mean any group/combination of three.

• #3

In poker, three of a kind is called «a set». In general usage, «a few» is often used to mean a group of 3 and «several» to mean a group of 4 or more. It’s not a rule or anything, just seems to be an adopted means among many to describe groups of things.

• #4

I am wondering if, in the same of «tupla» indicating a group of two things, there is a word to say a group of three. In Spanish, we have «tripleta» but I don’t have a clue if there is such a word in English.

There’s ‘trio’ which you might use for a group of musicians (‘string trio’), or three people (‘after their crime, the trio were arrested’). Google TRIO for more examples.

Does this help?

• #5

http://en.wikipedia.org/wiki/Tuple: In mathematics, a tuple is a finite sequence (also known as an «ordered list») of objects, each of a specified type. A tuple containing n objects is known as an «n-tuple». A 2-tuple is called a pair; a 3-tuple is a triple or triplet.

Not much useful outside a math class, probably.

Stef

• #6

You need to tell us what these elements are.

If musicians, you have a trio.

If musical notes, you may have a triad, or a triplet.

If children born of the same mother on the same day, you have triplets.

If pictures mounted adjacent to one another, you have a triptych.

Without more information any of the suggestions might be right, and all of them may be wrong.

• #7

Hi Everybody.

I am wondering if, in the same of «tupla» indicating a group of two things, there is a word to say a group of three. In Spanish, we have «tripleta» but I don’t have a clue if there is such a word in English.

Thanks a lot for your help
Cristinaalp

If they are a group of three actual elements from Earth, then it would be called a triad.

• #8

Trio is a counterpart of Duo or duet.
Triplet is a counterpart of Doublet.
Triplet is a counterpart of twin, if you are speaking of multiple births.
Threesome is a counterpart of Couple.
In some card games, «three of a kind» is the counterpart of «a pair.«
Triad and Trinity are other words that means a group of three, but I can’t think of counterparts for those.

• #9

Hi everybody.

Thanks a lot for all your replies and sorry about the vagueness of my post.
Actually, what I want to describe is a thing that is described by 3 elements (frequency, x, y).

So according to Stefan Ivanovich, I could say that the element is a 3-tuple that comprises (frequency, x,Y). Am I right?

Thanks a lot for all your help, Happy Holidays,
Cristina

• #10

2 cents
What I know is «duplet», but I’m not sure.

• #11

Hi everybody.

Thanks a lot for all your replies and sorry about the vagueness of my post.
Actually, what I want to describe is a thing that is described by 3 elements (frequency, x, y).

So according to Stefan Ivanovich, I could say that the element is a 3-tuple that comprises (frequency, x,Y). Am I right?

Thanks a lot for all your help, Happy Holidays,
Cristina

Yes your could say 3-tuple, but triplet would be better as implied by the end of my previous post (a 3-tuple is a triple or triplet). Just see the quoted wikipedia article for more.

Trying to help.

Stef

From Wikipedia, the free encyclopedia

In mathematics, especially in the area of abstract algebra known as combinatorial group theory, the word problem for a finitely generated group G is the algorithmic problem of deciding whether two words in the generators represent the same element. More precisely, if A is a finite set of generators for G then the word problem is the membership problem for the formal language of all words in A and a formal set of inverses that map to the identity under the natural map from the free monoid with involution on A to the group G. If B is another finite generating set for G, then the word problem over the generating set B is equivalent to the word problem over the generating set A. Thus one can speak unambiguously of the decidability of the word problem for the finitely generated group G.

The related but different uniform word problem for a class K of recursively presented groups is the algorithmic problem of deciding, given as input a presentation P for a group G in the class K and two words in the generators of G, whether the words represent the same element of G. Some authors require the class K to be definable by a recursively enumerable set of presentations.

History[edit]

Throughout the history of the subject, computations in groups have been carried out using various normal forms. These usually implicitly solve the word problem for the groups in question. In 1911 Max Dehn proposed that the word problem was an important area of study in its own right,^[1] together with the conjugacy problem and the group isomorphism problem. In 1912 he gave an algorithm that solves both the word and conjugacy problem for the fundamental groups of closed orientable two-dimensional manifolds of genus greater than or equal to 2.^[2] Subsequent authors have greatly extended Dehn’s algorithm and applied it to a wide range of group theoretic decision problems.^[3][4][5]

It was shown by Pyotr Novikov in 1955 that there exists a finitely presented group G such that the word problem for G is undecidable.^[6] It follows immediately that the uniform word problem is also undecidable. A different proof was obtained by William Boone in 1958.^[7]

The word problem was one of the first examples of an unsolvable problem to be found not in mathematical logic or the theory of algorithms, but in one of the central branches of classical mathematics, algebra. As a result of its unsolvability, several other problems in combinatorial group theory have been shown to be unsolvable as well.

It is important to realize that the word problem is in fact solvable for many groups G. For example, polycyclic groups have solvable word problems since the normal form of an arbitrary word in a polycyclic presentation is readily computable; other algorithms for groups may, in suitable circumstances, also solve the word problem, see the Todd–Coxeter algorithm^[8] and the Knuth–Bendix completion algorithm.^[9] On the other hand, the fact that a particular algorithm does not solve the word problem for a particular group does not show that the group has an unsolvable word problem. For instance Dehn’s algorithm does not solve the word problem for the fundamental group of the torus. However this group is the direct product of two infinite cyclic groups and so has a solvable word problem.

A more concrete description[edit]

In more concrete terms, the uniform word problem can be expressed as a rewriting question, for literal strings.^[10] For a presentation P of a group G, P will specify a certain number of generators

x, y, z, …

for G. We need to introduce one letter for x and another (for convenience) for the group element represented by x⁻¹. Call these letters (twice as many as the generators) the alphabet for our problem. Then each element in G is represented in some way by a product

abc … pqr

of symbols from , of some length, multiplied in G. The string of length 0 (null string) stands for the identity element e of G. The crux of the whole problem is to be able to recognise all the ways e can be represented, given some relations.

The effect of the relations in G is to make various such strings represent the same element of G. In fact the relations provide a list of strings that can be either introduced where we want, or cancelled out whenever we see them, without changing the ‘value’, i.e. the group element that is the result of the multiplication.

For a simple example, take the presentation {a | a³}. Writing A for the inverse of a, we have possible strings combining any number of the symbols a and A. Whenever we see aaa, or aA or Aa we may strike these out. We should also remember to strike out AAA; this says that since the cube of a is the identity element of G, so is the cube of the inverse of a. Under these conditions the word problem becomes easy. First reduce strings to the empty string, a, aa, A or AA. Then note that we may also multiply by aaa, so we can convert A to aa and convert AA to a. The result is that the word problem, here for the cyclic group of order three, is solvable.

This is not, however, the typical case. For the example, we have a canonical form available that reduces any string to one of length at most three, by decreasing the length monotonically. In general, it is not true that one can get a canonical form for the elements, by stepwise cancellation. One may have to use relations to expand a string many-fold, in order eventually to find a cancellation that brings the length right down.

The upshot is, in the worst case, that the relation between strings that says they are equal in G is an Undecidable problem.

Examples[edit]

The following groups have a solvable word problem:

• Automatic groups, including:
  • Finite groups
  • Negatively curved (aka. hyperbolic) groups
  • Euclidean groups
  • Coxeter groups
  • Braid groups
  • Geometrically finite groups
• Finitely generated free groups
• Finitely generated free abelian groups
• Polycyclic groups
• Finitely generated recursively absolutely presented groups,^[11] including:
  • Finitely presented simple groups.
• Finitely presented residually finite groups
• One relator groups^[12] (this is a theorem of Magnus), including:
  • Fundamental groups of closed orientable two-dimensional manifolds.
• Combable groups
• Autostackable groups

Examples with unsolvable word problems are also known:

• Given a recursively enumerable set A of positive integers that has insoluble membership problem, ⟨a,b,c,d | aⁿbaⁿ = cⁿdcⁿ : n ∈ A⟩ is a finitely generated group with a recursively enumerable presentation whose word problem is insoluble^[13]
• Every finitely generated group with a recursively enumerable presentation and insoluble word problem is a subgroup of a finitely presented group with insoluble word problem^[14]
• The number of relators in a finitely presented group with insoluble word problem may be as low as 14 ^[15] or even 12.^[16][17]
• An explicit example of a reasonable short presentation with insoluble word problem is given in Collins 1986:^[18][19]

Partial solution of the word problem[edit]

The word problem for a recursively presented group can be partially solved in the following sense:

Given a recursive presentation P = ⟨X|R⟩ for a group G, define:

then there is a partial recursive function f_P such that:

More informally, there is an algorithm that halts if u=v, but does not do so otherwise.

It follows that to solve the word problem for P it is sufficient to construct a recursive function g such that:

However u=v in G if and only if uv⁻¹=1 in G. It follows that to solve the word problem for P it is sufficient to construct a recursive function h such that:

Example[edit]

The following will be proved as an example of the use of this technique:

Theorem: A finitely presented residually finite group has solvable word problem.

Proof: Suppose G = ⟨X|R⟩ is a finitely presented, residually finite group.

Let S be the group of all permutations of N, the natural numbers, that fixes all but finitely many numbers then:

1. S is locally finite and contains a copy of every finite group.
2. The word problem in S is solvable by calculating products of permutations.
3. There is a recursive enumeration of all mappings of the finite set X into S.
4. Since G is residually finite, if w is a word in the generators X of G then w ≠ 1 in G if and only of some mapping of X into S induces a homomorphism such that w ≠ 1 in S.

Given these facts, algorithm defined by the following pseudocode:

For every mapping of X into S
    If every relator in R is satisfied in S
        If w ≠ 1 in S
            return 0
        End if
    End if
End for

defines a recursive function h such that:

This shows that G has solvable word problem.

Unsolvability of the uniform word problem[edit]

The criterion given above, for the solvability of the word problem in a single group, can be extended by a straightforward argument. This gives the following criterion for the uniform solvability of the word problem for a class of finitely presented groups:

To solve the uniform word problem for a class K of groups, it is sufficient to find a recursive function that takes a finite presentation P for a group G and a word in the generators of G, such that whenever G ∈ K:

Boone-Rogers Theorem: There is no uniform partial algorithm that solves the word problem in all finitely presented groups with solvable word problem.

In other words, the uniform word problem for the class of all finitely presented groups with solvable word problem is unsolvable. This has some interesting consequences. For instance, the Higman embedding theorem can be used to construct a group containing an isomorphic copy of every finitely presented group with solvable word problem. It seems natural to ask whether this group can have solvable word problem. But it is a consequence of the Boone-Rogers result that:

Corollary: There is no universal solvable word problem group. That is, if G is a finitely presented group that contains an isomorphic copy of every finitely presented group with solvable word problem, then G itself must have unsolvable word problem.

Remark: Suppose G = ⟨X|R⟩ is a finitely presented group with solvable word problem and H is a finite subset of G. Let H^* = ⟨H⟩, be the group generated by H. Then the word problem in H^* is solvable: given two words h, k in the generators H of H^*, write them as words in X and compare them using the solution to the word problem in G. It is easy to think that this demonstrates a uniform solution of the word problem for the class K (say) of finitely generated groups that can be embedded in G. If this were the case, the non-existence of a universal solvable word problem group would follow easily from Boone-Rogers. However, the solution just exhibited for the word problem for groups in K is not uniform. To see this, consider a group J = ⟨Y|T⟩ ∈ K; in order to use the above argument to solve the word problem in J, it is first necessary to exhibit a mapping e: Y → G that extends to an embedding e^*: J → G. If there were a recursive function that mapped (finitely generated) presentations of groups in K to embeddings into G, then a uniform solution of the word problem in K could indeed be constructed. But there is no reason, in general, to suppose that such a recursive function exists. However, it turns out that, using a more sophisticated argument, the word problem in J can be solved without using an embedding e: J → G. Instead an enumeration of homomorphisms is used, and since such an enumeration can be constructed uniformly, it results in a uniform solution to the word problem in K.

Proof that there is no universal solvable word problem group[edit]

Suppose G were a universal solvable word problem group. Given a finite presentation P = ⟨X|R⟩ of a group H, one can recursively enumerate all homomorphisms h: H → G by first enumerating all mappings h^†: X → G. Not all of these mappings extend to homomorphisms, but, since h^†(R) is finite, it is possible to distinguish between homomorphisms and non-homomorphisms, by using the solution to the word problem in G. «Weeding out» non-homomorphisms gives the required recursive enumeration: h₁, h₂, …, h_n, … .

If H has solvable word problem, then at least one of these homomorphisms must be an embedding. So given a word w in the generators of H:

Consider the algorithm described by the pseudocode:

Let n = 0
    Let repeatable = TRUE
        while (repeatable)
            increase n by 1
            if (solution to word problem in G reveals hn(w) ≠ 1 in G)
                Let repeatable = FALSE
output 0.

This describes a recursive function:

The function f clearly depends on the presentation P. Considering it to be a function of the two variables, a recursive function has been constructed that takes a finite presentation P for a group H and a word w in the generators of a group G, such that whenever G has soluble word problem:

But this uniformly solves the word problem for the class of all finitely presented groups with solvable word problem, contradicting Boone-Rogers. This contradiction proves G cannot exist.

Algebraic structure and the word problem[edit]

There are a number of results that relate solvability of the word problem and algebraic structure. The most significant of these is the Boone-Higman theorem:

A finitely presented group has solvable word problem if and only if it can be embedded in a simple group that can be embedded in a finitely presented group.

It is widely believed that it should be possible to do the construction so that the simple group itself is finitely presented. If so one would expect it to be difficult to prove as the mapping from presentations to simple groups would have to be non-recursive.

The following has been proved by Bernhard Neumann and Angus Macintyre:

A finitely presented group has solvable word problem if and only if it can be embedded in every algebraically closed group

What is remarkable about this is that the algebraically closed groups are so wild that none of them has a recursive presentation.

The oldest result relating algebraic structure to solvability of the word problem is Kuznetsov’s theorem:

A recursively presented simple group S has solvable word problem.

To prove this let ⟨X|R⟩ be a recursive presentation for S. Choose a ∈ S such that a ≠ 1 in S.

If w is a word on the generators X of S, then let:

There is a recursive function such that:

Write:

Then because the construction of f was uniform, this is a recursive function of two variables.

It follows that: is recursive. By construction:

Since S is a simple group, its only quotient groups are itself and the trivial group. Since a ≠ 1 in S, we see a = 1 in S_w if and only if S_w is trivial if and only if w ≠ 1 in S. Therefore:

The existence of such a function is sufficient to prove the word problem is solvable for S.

This proof does not prove the existence of a uniform algorithm for solving the word problem for this class of groups. The non-uniformity resides in choosing a non-trivial element of the simple group. There is no reason to suppose that there is a recursive function that maps a presentation of a simple groups to a non-trivial element of the group. However, in the case of a finitely presented group we know that not all the generators can be trivial (Any individual generator could be, of course). Using this fact it is possible to modify the proof to show:

The word problem is uniformly solvable for the class of finitely presented simple groups.

See also[edit]

• Combinatorics on words
• SQ-universal group
• Word problem (mathematics)
• Reachability problem
• Nested stack automata (have been used to solve the word problem for groups)

Notes[edit]

References[edit]

• Boone, W.W.; Cannonito, F.B.; Lyndon, Roger C. (1973). Word problems : decision problems and the Burnside problem in group theory. Studies in logic and the foundations of mathematics. Vol. 71. North-Holland. ISBN 9780720422719.
• Boone, W. W.; Higman, G. (1974). «An algebraic characterization of the solvability of the word problem». J. Austral. Math. Soc. 18: 41–53. doi:10.1017/s1446788700019108.
• Boone, W. W.; Rogers Jr, H. (1966). «On a problem of J. H. C. Whitehead and a problem of Alonzo Church». Math. Scand. 19: 185–192. doi:10.7146/math.scand.a-10808.
• Borisov, V. V. (1969), «Simple examples of groups with unsolvable word problem», Akademiya Nauk SSSR. Matematicheskie Zametki, 6: 521–532, ISSN 0025-567X, MR 0260851
• Collins, Donald J. (1969), «Word and conjugacy problems in groups with only a few defining relations», Zeitschrift für Mathematische Logik und Grundlagen der Mathematik, 15 (20–22): 305–324, doi:10.1002/malq.19690152001, MR 0263903
• Collins, Donald J. (1972), «On a group embedding theorem of V. V. Borisov», Bulletin of the London Mathematical Society, 4 (2): 145–147, doi:10.1112/blms/4.2.145, ISSN 0024-6093, MR 0314998
• Collins, Donald J. (1986), «A simple presentation of a group with unsolvable word problem», Illinois Journal of Mathematics, 30 (2): 230–234, doi:10.1215/ijm/1256044631, ISSN 0019-2082, MR 0840121
• Collins, Donald J.; Zieschang, H. (1990), Combinatorial group theory and fundamental groups, Springer-Verlag, p. 166, MR 1099152
• Dehn, Max (1911), «Über unendliche diskontinuierliche Gruppen», Mathematische Annalen, 71 (1): 116–144, doi:10.1007/BF01456932, ISSN 0025-5831, MR 1511645, S2CID 123478582
• Dehn, Max (1912), «Transformation der Kurven auf zweiseitigen Flächen», Mathematische Annalen, 72 (3): 413–421, doi:10.1007/BF01456725, ISSN 0025-5831, MR 1511705, S2CID 122988176
• Kuznetsov, A.V. (1958). «Algorithms as operations in algebraic systems». Izvestia Akad. Nauk SSSR Ser Mat. 13 (3): 81.
• Miller, C.F. (1991). «Decision problems for groups — survey and reflections». Algorithms and Classification in Combinatorial Group Theory. Mathematical Sciences Research Institute Publications. Vol. 23. Springer. pp. 1–60. doi:10.1007/978-1-4613-9730-4_1. ISBN 978-1-4613-9730-4.
• Nyberg-Brodda, Carl-Fredrik (2021), «The word problem for one-relation monoids: a survey», Semigroup Forum, 103 (2): 297–355, arXiv:2105.02853, doi:10.1007/s00233-021-10216-8
• Rotman, Joseph (1994), An introduction to the theory of groups, Springer-Verlag, ISBN 978-0-387-94285-8
• Stillwell, J. (1982). «The word problem and the isomorphism problem for groups». Bulletin of the AMS. 6: 33–56. doi:10.1090/s0273-0979-1982-14963-1.