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Universal algebra - roughly - is the study, construed broadly, of classes of algebraic structures (in a given language) defined by equations. Of course, it is really much more than that, but that's the best one-sentence definition I can give.

I'm very new to universal algebra. So far I've found it incredibly interesting, mainly because it looks at things I was already interested in from a new (to me) perspective, and that's always good; but I don't at all have a firm command of even the basics. For example, the recent question Relatively free algebras in a variety generated by a single algebra made me realize that I'd naively accepted a very false statement: until I thought about it, I'd sort of taken for granted that A is always relatively free in Var(A).

I'm sure this isn't the only false belief I have about universal algebra, and I'm sure I'll hold more in the future; and I'm also sure I'm not alone in this. So my question is:

What are some notable counterexamples (to reasonable hypotheses a student in universal algebra might have) in universal algebra?

I'm specifically interested in universal algebra because, well, it's fairly universal; it seems reasonable that a counterexample in universal algebra would be of interest to algebraists of many different stripes, and hopefully many outside algebra as well. At the same time, universal algebra is distinct enough that counterexamples in universal algebra would hopefully have their own flavor not found necessarily in questions like "counterexamples in group theory," "counterexamples in ring theory," etc. In that vein, I'd especially appreciate counterexamples about topics firmly within universal algebra - say, congruence lattices, or Mal'cev conditions - which nonetheless have "something to say" to other areas of mathematics.

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    $\begingroup$ Well, so far, Benjamin Steinberg seems to be your go-to man! $\endgroup$
    – Todd Trimble
    Commented Jan 9, 2014 at 4:17
  • $\begingroup$ I'm not sure this answers your question precisely, so I'm posting it as a comment instead: I always thought that the first isomorphism theorem only worked for some types of algebra, like groups and rings. It works for all universal algebras, including the algebras with no operations (i.e., sets). $\endgroup$
    – Pace Nielsen
    Commented Jan 11, 2022 at 5:33

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All free Jónsson-Tarski algebras on a finite nonempty set of generators are isomorphic. Thus free objects may not know their rank.

Curiously the automorphism group of this free algebra is the famous Thompson simple group $V$.

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  • $\begingroup$ Just checking: a Jonsson-Tarski algebra is an algebra in the language of a single binary function symbol $f$, with infinite domain $A$ such that $f$ is a bijection between $A^2$ and $A$, right? $\endgroup$
    – Noah Schweber
    Commented Jan 9, 2014 at 3:39
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    $\begingroup$ @NoahS I think you also want unary unpairing functions $g$ and $h$, so the identities are $g(f(x,y))=x,\ h(f(x,y))=y$, and $f(g(x),h(x))=x$. $\endgroup$
    – bof
    Commented Jan 9, 2014 at 3:58
  • $\begingroup$ In a variety generated by an infinite primal algebra $A$, all algebras freely generated by finitely many elements are isomorphic to the direct power $A^{A}$, and $\textrm{Aut}(A^{A})\simeq\textrm{Sym}(A)$. $\endgroup$
    – Joseph Van Name
    Commented Jan 9, 2014 at 14:47
  • $\begingroup$ Also, Higman proved that in the variety of $k$-Jónsson-Tarski algebras (= set $A$ with bijection $A^k\to A$, implemented in universal algebra by a $k$-ary law and $k$ unary laws) with free algebra $F_\ell^{(k)}$ on $\ell$ generators, for finite $\ell,\ell'\ge 1$ we have $F_\ell^{(k)}$ and $F_{\ell'}^{(k)}$ isomorphic iff $k-1$ divides $\ell-\ell'$. For instance $F_1^{(3)}$ is isomorphic to $F_3^{(3)}$ but not to $F_1^{(2)}$. $\endgroup$
    – YCor
    Commented Feb 17, 2020 at 14:01
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McKenzie proved that it is undecidable whether a finite universal algebra has a finite basis of identities.

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    $\begingroup$ That was actually part of a creative outburst of McKenzie's regarding decidability of issues such as residual finiteness, whether certain types (tame conrguence theory) occur in an algebra in the variety generated by a variety, and others. The counterexample was to the statement "These problems are not going to be solved in the millenium they were posed." Gerhard "Or In The Same Century, Anyway" Paseman, 2013.01.09 $\endgroup$
    – Gerhard Paseman
    Commented Jan 9, 2014 at 22:24
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Following George Bergman's recent preprint

http://arxiv.org/abs/1309.0564

(btw, he is very good at finding strange (counter-)examples in universal algebra!)

we recently found out that the universal group of the subsemigroup from the free monoid $\{a,b,c\}^*$ generated by $\{bc,abcabc,bcabca,bcabcabcabc\}$ is isomorphic to $(\mathbb{Z}\times\mathbb{Z})\ast\mathbb{Z}$, and so non-free, which somehow collides with Nielsen-Schreier theorem.

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    $\begingroup$ So Nielsen-Schreier theorem is false ? $\endgroup$
    – Joël
    Commented Jan 9, 2014 at 15:51
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    $\begingroup$ @Jöel: $({\Bbb{Z}}\times {\Bbb{Z}})*{\Bbb{Z}}$ is free product, but isn't a free group. $\endgroup$
    – janmarqz
    Commented Jan 11, 2014 at 0:57
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    $\begingroup$ Okay, I get it: the analog of Nielsen-Schreier for monoids is false. Sorry for having been thick. $\endgroup$
    – Joël
    Commented Jan 11, 2014 at 1:30
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    $\begingroup$ @Joël: well, before this example, i was very sure that universal groups of subsemigroups of free monoids must be free (subsemigroups of free monoids are of course not necessarily free), since subsemigroups of free monoids live inside free groups and kind of their universal groups would be some subgroups from the free group. Well, semigroups are just crazy wild objects! $\endgroup$
    – Victor
    Commented Jan 12, 2014 at 7:04
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There are compact totally disconnected lattices which are not inverse limits of finite lattices.

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Subalgebras of free algebras are free. True for groups and Lie algebras but false for semigroups or commutative rings.

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    $\begingroup$ Rather blatantly false for lattices. $\endgroup$
    – bof
    Commented Jan 9, 2014 at 3:53
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    $\begingroup$ False also for Boolean algebras since the finite free Boolean algebras have cardinality of the form $2^{2^{n}}$. $\endgroup$
    – Joseph Van Name
    Commented Jan 9, 2014 at 5:10
  • $\begingroup$ @JosephVanName Oh, right. And I believe every countable Boolean algebra is embeddable in the free Boolean algebra on $\aleph_0$ generators? $\endgroup$
    – bof
    Commented Jan 9, 2014 at 5:28
  • $\begingroup$ Also false for "most" varieties of groups: it only holds for the variety of all groups, the variety of all abelian groups, the variety of all abelian groups of exponent $p$ ($p$ a prime), and the trivial variety. $\endgroup$
    – Arturo Magidin
    Commented Jan 10, 2014 at 4:54
  • $\begingroup$ Another example: $R$-modules whenever $R$ is a commutative ring that isn't a PID. $\endgroup$
    – goblin GONE
    Commented Jun 14, 2016 at 17:30
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A deep theorem of Oates and Powell shows that any finite group has a finite basis for its identities. One might think that the same is true for semigroups. But Perkins showed the 6-element semigroup consisting of the $2\times 2$ identity matrix, the zero matrix and the four matrix units $E_{ij}$ is not finitely based.

Mark Sapir, in a tour-de-force work involving symbolic dynamics, proved that this semigroup is inherently nonfinitely based, meaning it cannot belong to any finitely based locally finite variety. Hence any finite semigroup generating a variety containing this semigroup is not finitely based.

See Sapir's book http://www.math.vanderbilt.edu/~msapir/book/b2.pdf for this and many other nice universal algebra results.

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    $\begingroup$ Benjamin, can you add a reference for Mark Sapir's tour-de-force? $\endgroup$
    – Todd Trimble
    Commented Jan 9, 2014 at 4:18
  • $\begingroup$ You can find it in in his book math.vanderbilt.edu/~msapir/book/b2.pdf where in Chapter 3 he uses symbolic dynamics to characterize in an algorithmic way all finite inherently non finitely based semigroups. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 9, 2014 at 14:53
  • $\begingroup$ Bryant (1982) proved that the theorem of Oates and Powell does not generalize to "pointed" groups. That is, if you take a finite group and simply identify one of its elements as "special," then it no longer has a finite basis for its identities. That seems striking at first, but think about all the new identities you have once you can recognize a special element of the group (besides the identity, of course). $\endgroup$
    – William DeMeo
    Commented Feb 22, 2017 at 7:45
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A complete semilattice is automatically also a complete lattice. Hence a student might expect that there is no need to distinguish between complete semilattices and complete lattices. However, the complete homomorphisms of complete meet-semilattices in general don't preserve the join operation, so the two structures must really be distinguished.

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Here is another great Mark Sapir result. Let $S$ be the three element cyclic semigroup $\langle x\mid x^2=x^3\rangle$. Then $S$ has no finite basis for its quasi-identities. I believe Jackson and Volkov later showed that any finite semigroup containing this one also has no finite basis for its quasi-identities.

Related, Mark Sapir showed that although the variety generated by the finite semigroup $\{1,a,b\}$ where $1$ is the identity and $xy=x$ for $x,y\neq 1$ has only fnitely many subvarieties, it has uncountably many subquasivarieties. I recently showed with Margolis and Saliola that $\{1,a,b\}$ has no finite basis of quasi-identities (this is easier than Sapir's result) using hyperplane arrangements.

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    $\begingroup$ $\langle x\;|\;x^2=x^3\rangle$ should be $\langle x\;|\;x^3=x^4\rangle$. $\endgroup$
    – Keith Kearnes
    Commented Mar 25 at 16:20
  • $\begingroup$ @KeithKearnes you are correct $\endgroup$
    – Benjamin Steinberg
    Commented Mar 25 at 17:48
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Here is another one I love. Marcel Jackson showed that there is a finite semigroup with a finite basis for its identities such that the variety it generates contains uncountably many subvarieties. I would have thought this impossible. Here is the link:

http://www.sciencedirect.com/science/article/pii/S0021869399982807

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Not sure whether you count this as universal algebra; someone classically-minded probably wouldn't, and someone categorically-minded probably would.

In any case: it's possible to cook up two nonisomorphic operads that give rise to the same algebraic theory. Here "operad" means "non-symmetric operad of sets", and by "give rise to" I'm referring to the fact that every operad $P$ has an associated algebraic theory (which expressed as a monad is $\coprod_{n \geq 0} P_n \times (-)^n$).

This is a counterexample to the general belief that operads are algebraic theories of a special kind. You still see this belief expressed all over the place, and it's reinforced by the fact that for symmetric operads, different operads genuinely do give rise to different theories.

Edit Here are some details. From any operad $P$, we can construct its reverse $P^\#$. It has the same operations as $P$, and the same identity operation, but the order of the composition is reversed. Thus, $\theta \circ (\theta_1, \ldots, \theta_n)$ in $P^\#$ is $\theta \circ (\theta_n, \ldots, \theta_1)$ in $P$. It's easy to show that $P$ and $P^\#$ give rise to the same algebraic theory.

So, to construct our counterexample, we just have to find some operad not isomorphic to its reverse. This is a bit harder than it sounds. For instance, any operad that admits a symmetric structure is isomorphic to its reverse, and several other well-known operads are too. But you can find one in arXiv:math/0404016. It's obscure enough that I couldn't immediately remember what it was. There may be simpler examples out there; I think Steve Lack told me one by email.

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    $\begingroup$ Tom, could you give a reference for this, or maybe just give two such operads in your answer if it's not too much trouble? $\endgroup$
    – Todd Trimble
    Commented Jan 9, 2014 at 20:55
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This is only a counterexample in the loose sense of the word. Universal algebra students may hypothesize that all interesting fundamental operations studied in universal algebra and related areas have arity at most 2. Even though ternary terms are useful when investigating certain properties of congruence lattices (i.e. Mal'cev conditions), ternary terms rarely pop up as fundamental operations on algebras.. Therefore, algebras with interesting ternary operations are helpful to keep in mind. For example, median algebras are algebras with a single ternary operation $m$ that satisfies the identities: $$m(a,a,b)=a$$ $$m(a,b,c)=m(b,a,c)=m(b,c,a)$$ $$m(m(a,b,c),d,e)=m(a,m(b,d,e),m(c,d,e)).$$

For example, if $X$ is a distributive lattice, then $(X,m)$ is a median algebra where $$m(x,y,z)=(x\wedge y)\vee(x\wedge z)\vee(y\wedge z)=(x\vee y)\wedge(x\vee z)\wedge(y\vee z).$$

See this question for more ternary operations in universal algebra.

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    $\begingroup$ Median algebras are essentially the same thing as Cat(0) cube complexes so are fundamental to 3-manifold topology. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 9, 2014 at 14:56
  • $\begingroup$ @BenjaminSteinberg That sounds interesting. Do you know a reference for the connection? $\endgroup$
    – arsmath
    Commented Feb 9, 2015 at 14:34
  • $\begingroup$ @arsmath, look at personal.soton.ac.uk/gan/Roller.pdf $\endgroup$
    – Benjamin Steinberg
    Commented Feb 9, 2015 at 16:30
  • $\begingroup$ @BenjaminSteinberg Thanks! That's fascinating reading. $\endgroup$
    – arsmath
    Commented Feb 10, 2015 at 21:15
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When one first learns about Birkhoff's HSP Theorem and the variety $\mathcal{V}(\mathbf A)$ generated by the algebra $\mathbf A$, it's a common mistake to think that properties of $\mathbf A$ are inherited by all inhabitants of $\mathcal{V}(\mathbf A)$. For example, if $\mathbf A$ has a modular congruence lattice, then so should every algebra in $\mathcal{V}(\mathbf A)$. (false)

A rule of thumb is to consider whether the property in question is "verbal"---that is, whether it can be specified by a set of equations. If so, then the property is inherited by all inhabitants of the variety.

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There are several very basic facts coming from traditional algebra that are confronted when going to a more "universal" setting. I guess that students that know some group/ring theory might be confused by the fact that

kernels of homomorphisms (in general, congruences and quotients) are not determined by the preimage of $0$.

Although congruences $\theta$ are still determined by $0/\theta$ for Boolean algebras, this doesn't hold for bounded lattices.

When starting studying some lattice theory, it might be difficult to realize that

a subposet of a lattice $L$ that is indeed a lattice need not be a sublattice of $L$.

Finally, I conclude with a horrible example concerning factorization of structures that is somehow related to the previous example. Consider the direct product of two finite, totally ordered semilattices, as in the picture.

direct product of semilattice chains

Actually, this product has a lattice order. If you take the subset consisting of the filled points, it is a subsemilattice, it is lattice-ordered and it's still the product of two (smaller) chains.

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Let me mention the question we struggled with for some time: we could not find examples showing that if you take away finitely many elements from a finitely generated semigroup, so that the resulting subset is a subsemigroup, then neither hopficity, not co-copficity is preserved in general. Strangely, but it's exactly relating (co-)hopficity for semigroups to (co-)hopficity for graphs what made it possible. Here are two links:

http://arxiv.org/abs/1305.6176

http://arxiv.org/abs/1307.6929

(I have to acknowledge though that Ben Steinberg finds Rees index unnatural)

Some other surprising links of semigroup properties to algebraic properties of various relational structures can be found in works on FA-presentable semigroups.

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Regarding a small model of High School Algebra

M. Jackson, A note on HSI-algebras and counterexamples to Wilkie's identity, Algebra Universalis 36 (1996), 528–535.

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  1. May be this is too naive, but for people accustomed to more or less conventional algebraic objects (like groups, Lie and associative algebras, etc.), it could be revealing to know that there are universal algebras without proper subalgebras (for one such class of algebars, see, e.g., K. Kaarli, Subalgebras of the squares of weakly diagonal majority algebras, Studia Sci. Math. Hungar. 49 (2012), N4, 509-524)

  2. Free semigroups can be embedded into metabelian groups (see, e.g. G. Bergman, Hyperidentities of groups and semigroups, Aequat. Math. 22 (1981), 315-317).

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Not all nonabelian relatively free groups of exponent zero or a prime power are directly indecomposable; not all splitting groups in a variety of exponent zero or a prime power are relatively free.

These were Problems 21 and 22 in Hanna Neumann's Varieties of Groups; Peter Neumann provided examples in A note on the direct decomposablity of relatively free groups. Quart. J. Math. Oxford Ser. (2) 19 1968 67–79, MR 0223437 (36 #6485)

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For a while it was conjectured that modularity was the "weakest" congruence identity in the following sense: if there was some non-trivial identity satisfied by the congruence lattice of every algebra in a variety, then the variety would have to be congruence modular. This was refuted by Polin in 1977.

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My favorite "counterexample" in universal algebra is the one from Algebras, Lattices, Varieties (Alvin), Exercise 5.1.

Consider two finite algebras that are the same except for a relabeling of the operations. Our intuition tells us (rightly, imho) that these are simply different ways if presenting the same mathematical structure. Formally, however, not only are these distinct mathematical objects, but also they are not even isomorphic!

You can find more details in Exercise 5.1 of the Alvin book mentioned above, or see this blog post.

Fortunately, there are other notions of equivalence (like "term-equivalence" and "polynomial-equivalence") that align better with our intuition in such contexts.

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    $\begingroup$ Does your intuition really tell you that? Switching addition and multiplication in a ring $R$ is such a bizarre thing to do that my intuition tells me that there's nothing good coming out of it. In particular, it's immediately clear that for $R\neq 0$ the resulting structure cannot be isomorphic to the original (or any ring), because $(R,+)$ is a group while $(R,\cdot)$ isn't (or alternatively because distribution holds only in one way not the other). $\endgroup$
    – Johannes Hahn
    Commented Feb 25, 2017 at 10:50
  • $\begingroup$ @JohannesHahn Consider this example $\mathbf A = \langle \{0,1\}, f^{\mathbf A}, g^{\mathbf A}\rangle$ where $f^{\mathbf A}$ is the identity on $\{0,1\}$ and $g^{\mathbf A}(0) = 1$ and $g^{\mathbf A}(1) = 0$. Let $\mathbf B = \langle \{0,1\}, f^{\mathbf B}, g^{\mathbf B}\rangle$ where $f^{\mathbf B}(0) = 1$ and $f^{\mathbf B}(1) = 0$ and $g^{\mathbf B}$ is the identity on $\{0,1\}$. Of course, $\mathbf A$ and $\mathbf B$ are syntactically distinct algebraic structures, and they are nonisomorphic, but if we treated them as "nonindexed" algebras, they would be the same. $\endgroup$
    – William DeMeo
    Commented Feb 25, 2017 at 11:03
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    $\begingroup$ Exactly, they are syntactically different. And that syntax difference is there for a semantic reason: They simply mean completely different things (like addition and multiplication). For me it's totally non-intuitive to even think about such an reordering. Why would anyone want to do that? The only reason I can think of is accidental like with $K$-vector spaces where we really only have one operation $K\times V\to V$ besides addition but for formal reasons define $|K|$-many operations $V\to V$ instead. And these can be reasonably reordered because $K$ might have non-trivial automorphisms. $\endgroup$
    – Johannes Hahn
    Commented Feb 25, 2017 at 11:14
  • $\begingroup$ @JohannesHahn I completely disagree. It is non-intuitive to me that two structures that represent the same mathematical object, and are indistinguishable except for the order of their operations, should be considered not the same (or 'non-isomorphic"). $\endgroup$
    – William DeMeo
    Commented Jul 8, 2020 at 9:28
  • $\begingroup$ @JohannesHahn such objects are clearly semantically the same. They differ only in syntax. That is, therefore, not a good definition of "sameness." $\endgroup$
    – William DeMeo
    Commented Jul 8, 2020 at 9:29

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