# Counterexamples in algebra?

This is certainly related to "What are your favorite instructional counterexamples?", but I thought I would ask a more focused question. We've all seen Counterexamples in analysis and Counterexamples in topology, so I think it's time for: Counterexamples in algebra.

Now, algebra is quite broad, and I'm new at this, so if I need to narrow this then I will- just let me know. At the moment I'm looking for counterexamples in all areas of algebra: finite groups, representation theory, homological algebra, Galois theory, Lie groups and Lie algebras, etc. This might be too much, so a moderator can change that.

These counterexamples can illuminate a definition (e.g. a projective module that is not free), illustrate the importance of a condition in a theorem (e.g. non-locally compact group that does not admit a Haar measure), or provide a useful counterexample for a variety of possible conjectures (I don't have an algebraic example, but something analogous to the Cantor set in analysis). I look forward to your responses!

You can also add your counter-examples to this nLab page: http://ncatlab.org/nlab/show/counterexamples+in+algebra

(the link to that page is currently "below the fold" in the comment list so I (Andrew Stacey) have added it to the main question)

• My feeling is that this question is far too broad. Jun 21 '10 at 23:11
• I like that the question is general. I think if it's narrowed too much we won't get as many interesting responses. All of the big list type questions that have been successful have been fairly general, so I don't think it hurts as long as we aren't swarmed with questions like this. Jun 22 '10 at 0:23
• Meta discussion: tea.mathoverflow.net/discussion/459/counterexamples-in-algebra Jun 22 '10 at 7:54
• Whilst I like lists of counterexamples, I don't think that MO is an appropriate place for one. I've explained why in the meta discussion (NB: please vote for the comment linking to the meta discussion so that it appears "above the fold"). I think that this would work so much better as a wiki page. So I've started one on the nLab: ncatlab.org/nlab/show/counterexamples+in+algebra Obviously, as I'm not an algebraist I didn't understand everything and have probably left out a lot of information in copying it over. I recommend closing this question and redirecting to that nLab page. Jun 22 '10 at 8:33
• Andrew, why not keep the question open, in order to generate the examples here that can then be more sensibly organized on your page? It seems likely to me that you will get a lot of good examples with this question that might otherwise be missed. Jun 22 '10 at 13:16

## 58 Answers

In the category of rings, epimorphisms do not have to be surjective: $\mathbb{Z}\hookrightarrow \mathbb{Q}$.

• For more on epimorphisms that are not surjective you can take a look at the article of G.A.Reid "Epimorphisms and surjectivity" (Invent.9, 295-307, (1970)). He discussed this problem in the context of groups, $C^*$ algebras, von-Neumann algebras, (finite dim.) Lie algebras, semisimple Lie algebras, (locally) compact groups. For Hopf algebras you can take a look at the more recent paper arXiv:0907.2881. Jun 25 '10 at 8:05
• omg what's an epimorphism then? Oct 9 '10 at 22:26
• $f: A \to B$ is an epimorphism in any category if for every $g,h: B \to C$ $gf=hf$ implies that g=h. Oct 10 '10 at 17:34
• mathoverflow.net/questions/109/… Jun 18 '11 at 9:01
• @Nikita Kalinin: As you have probably noticed already, in your ring $S$ we have $x = 2xy = y$. Feb 27 '13 at 15:45

I like Lance Small's example of a right but not left Notherian ring: matrices of the form $\begin{pmatrix}a & b\\ 0 & c\end{pmatrix}$ where $a\in\mathbb{Z}$ and $b,c\in\mathbb{Q}$.

• Indeed, this is also an example that shows right global dimension is not equal to left global dimension. The former is 1 while the latter is at least 2 (and exactly 2 IIRC) Jun 18 '11 at 18:30

The ring $A = \prod_{n=1}^{\infty} \mathbb{F}_2$ has some interesting/disturbing properties.

For example, the affine scheme $X := {\rm{Spec}}(A)$ has non-open connected components (since it has infinitely many open points), all local rings on $X$ are noetherian (in fact they're all $\mathbb{F}_2$ since $a^2 = a$ for all elements $a$) even though $A$ is not noetherian, and if $I$ is an ideal that isn't finitely generated then ${\rm{Spec}}(A/I) \hookrightarrow X$ is formally unramified (since closed immersion), finite type, and flat but not étale (since not finitely presented) and not open, in contrast with the noetherian case.

• +1. I just want to add: $X$ is the Stone-Cech-compactification of the natural numbers. The structure sheaf is the constant sheaf $\mathbb{F}_2$. Jun 22 '10 at 10:52
• Wow,you're right,Martin-I didn't think of it that way! Jun 23 '10 at 0:45
• $A$ has a proper ideal $I=\oplus_{n=1}^\infty \mathbb F_2$. Since the quotient ring $A/I$ has unit element, Zorn says it has a maximal ideal, but one cannot explicitly produce a maximal ideal. Jun 28 '10 at 8:08
• @AAP: Yes, such maximal ideals correspond to free ultrafilters on the natural numbers. Jun 28 '10 at 10:37
• @unknown: unramified = formally unramified + locally of finite presentation. Jun 18 '11 at 9:01

A non-abelian group, all of whose subgroups are normal: the quaternion group, $$Q=\langle\thinspace a,b\thinspace|\thinspace a^4=1,a^2=b^2,ab=ba^3\thinspace\rangle$$

1) (Nagata) There are noetherian domains of infinte Krull dimension: Localize $k[x_1,x_2,...]$ at the prime ideals $(x_1),(x_2,x_3),(x_4,x_5,x_6),...$.

2) (Malcev) Every commutative cancellative monoid embeds into a group. This is false in the non-commutative case. A very instructive counterexample is given by $\langle a,b,c,d,x,y,u,v : ax=by, cx=dy, au=bv \rangle$.

3) The Theorem of Cantor-Bernstein for sets does not carry over to algebraic structures. For example, the fields $K=\overline{\mathbb{Q}(x_1,x_2,...)}$ (or $K=\mathbb{C}$) and $K(t)$ embed into each other, but they are not isomorphic.

The group $\mathbb{Z}^4$ is not the fundamental group of any 3-manifold, proved by Stallings in this 1962 paper. It follows that there is no algorithm for recognizing 3-manifold groups.

• Leave it to John to not only find the counterexample no one else thinks of,but to have the exact reference handy.We are very priviledged to have you as a contributor,Professor Stillwell. Keep up the awesome work. Jun 22 '10 at 2:41
• Algorithm in which sense? There's no algorithm to determine whether a presentation defines the trivial group (hence for recognizing 0-manifold groups)...
– YCor
Oct 31 '16 at 4:40
• @YCor This is the usual sense of algorithm. The point is that the algorithm has to recognize whether a given finite presentation is of a 3-manifold group. Contrast this with the problem of recognizing 4-manifold groups, which is trivially solvable because any finitely-presented group is the group of a 4-manifold. As you point out, the corresponding problem for 0-manifold groups is clearly not solvable. Nov 1 '16 at 18:52
• @JohnStillwell so I should just point out that the 3-dim case is an easy consequence of the 0-dimensional case and the fact that, for instance, 3-dimensional groups don't contain $\mathbf{Z}^4$. Say, because $(G*G)^4$ is a 3-dimensional group iff $G$ is trivial.
– YCor
Nov 1 '16 at 20:32

An exact sequence that does not split: $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$, where the first map is multiplication by 2.

• Does anyone think all exact sequences split? Jun 22 '10 at 1:22
• No, but I'd say this is a perfectly valid example of "illuminating a definition". The point isn't dispelling false beliefs, but clarifying concepts for people who are just learning the topic. Jun 22 '10 at 9:47
• @Steven: since I taught you about exact sequences, I'm certainly glad that you don't! More seriously, if someone met exact sequences of vector spaces first, they might have that misconception. Jun 23 '10 at 14:02
• You would think that all people doing spectral sequence calculations have all but forgotten this fact... no one bothers to write down extension problems anymore. Dec 22 '16 at 4:32

A number ring which is a principal ideal domain (and, hence, a unique factorization domain) but is not Euclidean: the ring of integers of ${\bf Q}(\sqrt{-19})$. See Th Motzkin, The Euclidean algorithm, Bull Amer Math Soc 55 (1949) 1142-1146, available at http://projecteuclid.org/euclid.bams/1183514381

• Does $$R[x] \cong S[x]$$ imply $$R \cong S$$? ( Taken from this link. )

• Here is a counterexample. Let $$R=\displaystyle\frac{\mathbb{C}[x,y,z]}{(xy-(1-z^2))}, \quad \ S= \displaystyle\frac{\mathbb{C}[x,y,z]}{(x^2y-(1-z^2))}$$ Then, $$R$$ is not isomorphic to $$S$$ but, $$R[T]\cong S[T]$$. In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3)

• http://www.math.lsa.umich.edu/~hochster/Lip.text.pdf

Excellent Counterexamples.

Let $$G$$ be a group and let $$\mathscr{S}(G)$$ denote the group of Inner-Automorphisms of $$G$$.

The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $$Z(G)$$ is the center of the group. Now, if $$Z(G) =\{e\}$$ then one can see that $$G \cong \mathscr{S}(G)$$. What about the converse? That is if $$G \cong \mathscr{S}(G)$$ does it imply that $$Z(G)=\{e\}$$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. That is if $$G \cong \mathscr{S}(G)$$ does it imply that $$Z(G)= \{e\}$$?

The answer is yes there are groups with non-trivial center which are isomorphic to $$\mathscr{S}(G)$$. The answer is given at this link

Next one:

• Does there exists a finite group $$G$$ and a normal subgroup $$H$$ of $$G$$ such that $$|Aut(H)|>|Aut(G)|$$

Arturo Magidin posed this question some time ago at MATH.SE

• Question. Can we have a finite group $$G$$, normal subgroups $$H$$ and $$K$$ that are isomorphic as groups, $$G/H$$ isomorphic to $$G/K$$, but no $$\varphi\in\mathrm{Aut}(G)$$ such that $$\varphi(H) = K$$?

• Answer was provided by Vipul Naik. Link is given here.

Question was posed by Zev Chonoles at $$\textbf{MATH.SE}$$

• I know it is possible for a group $$G$$ to have normal subgroups $$H, K$$, such that $$H\cong K$$ but $$G/H\not\cong G/K$$, but I couldn't think of any examples with $$G$$ finite. What is an illustrative example?

• Answer from this link: Take $$G = \mathbb{Z}_4 \times \mathbb{Z}_2$$, $$H$$ generated by $$(0,1)$$, $$K$$ generated by $$(2,0)$$. Then $$H \cong K \cong \mathbb{Z}_2$$ but $$G/H \cong \mathbb{}Z_4$$ while $$G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$.

• Math.se is coincidentally an actual Swedish math website also... math.se Oct 22 '15 at 21:15

A number field where the ring of integers is Euclidean but not norm-Euclidean: ${\bf Q}(\sqrt{69})$. See David A Clark, A quadratic field which is Euclidean but not norm-Euclidean, Manuscripta Mathematica 83 (1994) 327-330.

In group theory, Lagrange's Theorem states that the order of a subgroup divides the order of the group, however the converse is false. The usual counterexample given is the alternating group $A_4$ of order 12 which has no subgroup of order 6.

• Having internalised this counterexample in my undergraduate days, I was completely gobsmacked by Hall's theorem in my first-year algebra class. Jun 19 '11 at 16:47
• Also, Lagrange's Theorem for groups does not hold for loops (does not hold without associativity). Aug 9 '15 at 13:27
• The higher alternating groups will work as well. More generally, any nonabelian simple group must have even order but cannot have a subgroup of index 2, for it would automatically be normal. Mar 22 '18 at 7:37

From Milnor's book "Algebraic K-Theory":

A (nonzero!) associative ring for which a free module of rank 2 is isomorphic to a free module of rank 1: The ring of endomorphisms of an infinite-dimensional vector space.

Two finite non-isomorphic groups with the same order profile: let $C_n$ be the cyclic group of $n$ elements, let $Q=\langle\thinspace a,b\thinspace|\thinspace a^4=1,a^2=b^2,ab=ba^3\thinspace\rangle$ be the quaternion group, then $C_4\times C_4$ and $C_2\times Q$ are not isomorphic (the first is abelian, the second is not) but both have 1 element of order 1, 3 elements of order 2, and 12 elements of order 4.

By contrast, if two finite abelian groups have the same order profile, then they are isomorphic.

• Such two groups are called conformal. See this MO question : mathoverflow.net/q/39848 . An easy example is that of $C_3\times C_3\times C_3$ and the group of matrices $I_3+N$ with nilpotent, over the field ${\mathbb F}_3$. Jan 23 '17 at 13:26

The ring $R = k[x,y]/(x^2, xy)$ is a simple example of a local commutative noetherian ring that is not Cohen-Macaulay. It is sometimes referred to as the "Emmy Ring."

This ring is very useful for showing how unintuitive non-CM rings can be. For instance, letting $I = (x)$, then $\operatorname{depth} R/I = 1 > 0 = \operatorname{depth} R$; in particular the (innocuous looking) inequality

$\operatorname{depth} R/I + \operatorname{grade} I \leq \operatorname{depth} R$

need not hold. Here $\operatorname{grade} I$ is the length of the longest regular sequence in $I$.

• The ring $R$ is not local. Dec 1 '15 at 8:01

Tarski's monsters: infinite groups in which every proper non-trivial subgroup is of prime order $p$. They are $2$-generated simple groups.

They were constructed by Olshanskii and as far as I remember they were also constructed independently by Rips, maybe even before Olshanskii, but he did not bother publishing it. Can anyone confirm this?

• Rips had some preliminary text published and also gave a series of talks. It is hard to tell if these ideas would lead to the construction of an actual example. Some key components of Olshanskii's construction are missing in Rips' constructions.
– user6976
Oct 9 '10 at 22:07
• I remember Rips talking about this in Oxford mid-late 70s and everyone then believed that he had constructed Tarski monsters. I was a graduate student then and do not know what Rips published. May 14 '14 at 19:28

Two famous cases that come to mind are:

1. Nagata's counterexample to Hilbert's fourteenth problem.

2. Counterexamples by various people to (the original version of) the Burnside problem.

• Can you give a brief description of Nagata's actual construction? Or a reference? (The wikipedia page doesn't say much). Jun 22 '10 at 1:26
• This is not my area at all, but this looks like a readable reference: math.bas.bg/serdica/2001/2001-171-192.pdf Jun 22 '10 at 2:18
• And Nagata's example of a Noetherian ring with infinite Krull dimension, as well as a noetherian domain whose normalization is not a module-finite extension. Jun 22 '10 at 2:27

A finite group in which a product of two commutators need not be a commutator: This is Exercise 3.27 in Rotman, The Theory of Groups, a construction attributed to Carmichael. Let $G$ be the subgroup of $S_{16}$ generated by the eight permutations $(ac)(bd)$, $(eg)(fh)$, $(ik)(jl)$, $(mo)(np)$, $(ac)(eg)(ik)$, $(ab)(cd)(mo)$, $(ef)(gh)(mn)(op)$, and $(ij)(kl)$. Then the commutator subgroup of $G$ is generated by the first four of these elements, and has order 16. It contains $\alpha=(ik)(jl)(mo)(np)$, but $\alpha$ is not a commutator.

Rotman remarks elsewhere that the smallest group in which there is a product of commutators which is not a commutator is a group of order 96.

• Leave it to Rotman to find such a cool counterexample and not hesitate to put it in one of his textbooks.This is why most of them are classics. Jun 22 '10 at 5:46
• While doing work at an REU some years ago, I needed to know that every element of $A_n$ was in fact a commutator. So I learned of this subtlety at a relatively early age -- it would certainly have made things easier if every element of the commutator subgroup was automatically a commutator. Jun 27 '10 at 23:16
• (But it is true for alternating groups, as I did not quite mention in my previous response. It seems that it's actually true for lots of groups, just not all of them...) Jun 27 '10 at 23:16
• Yes, Ore's conjecture, now proved by Liebeck, Shalev et al, says that in any finite non-Abelian simple group, every element is a commutator. This is not generally true even in perfect groups. May 17 '14 at 20:28

Grigorchuk 1984 example of a finitely generated group with intermediate growth (there are no such linear group).

An infinite group with exactly two conjugacy classes. See G. Higman, B. H. Neumann, and H. Neumann, Embedding theorems for groups, J. London Math. Soc. 24 (1949), 247-254.

• That is not a big deal. A much bigger deal is an infinite finitely generated group with 2 conjugacy classes constructed by Osin (Annals of Math, this year).
– user6976
Oct 9 '10 at 22:15
• Maybe it is not a big deal now, but note that the paper was published in 1949. May 14 '14 at 19:30

Higman's group $G=\left< a_1,\ldots, a_4 | \forall i\in\mathbb{Z}/4\mathbb{Z}: a_i=[a_{i+1},a_i] \right>$, which has no subgroups of finite index. See: G. Higman, A finitely generated infinite simple group, J. London Math. Soc. 26 (1951), 61-64.

There are finitely presented groups whose word problem is undecidable in computability theory.

An infinitely generated and non-Noetherian subring of a polynomial ring:

$$R=K[x,xy,xy^2,\ldots, xy^n,\ldots] \subset S=K[x,y].$$

Explanation The ring $R$ is graded and monomial: it is spanned by the monomials $x^ay^b$ that are contained in it, whose exponents are the lattice points in the cone $C=\{(a,b)=(0,0)$ or $a>0, b\geq 0\}.$ The minimal generators of the homogeneous ideal $R_{+}$ of positive degree elements correspond to the minimal generators $(1,n), n\geq 0$ of the lattice cone $C\cap\mathbb{Z}^2.$ Thus $R_{+}$ (respectively, $R$) is infinitely-generated with ideal (respectively, $K$-algebra) minimal generators $x,xy,xy^2,\ldots, xy^n,\ldots.$

• This idea can be used to construct a variety with a non-finitely generated ring of global sections. Jun 21 '11 at 21:24

I think the Prüfer group $G=\mathbb{Z}_{p^\infty}$ deserves its place here.

For example,

• it has two nonisomorphic subgroups $H_1, H_2$ such that $G/H_1 \simeq G/H_2$,
• moreover, it has a proper subgroup $0 \neq H \subseteq G$ such that $G \simeq G/H$,
• it is an infinite group whose proper subgroups are all finite,
• moreover, it is a non-cyclic group whose all proper subgroups are cyclic,
• it is Artinian and not Noetherian (as a $\mathbb{Z}$-module), ...

Harry Hutchins "Examples of commutative rings" may be of interest. It is based on his 1978 Chicago Ph.D. thesis under Kaplansky, and not surprisingly it serves as a useful complement to Kaplansky's excellent textbook Commutative Rings (most references to proofs refer to Kaplansky). There is also a 3 page list of errata, updates,... dated July 1983, which is distributed with the book.

Hutchins, Harry C. 83a:13001 13-02
Examples of commutative rings. (English)
Polygonal Publ. House, Washington, N. J., 1981. vii+167 pp. $13.75. ISBN 0-936428-05-8 The book is divided into two parts: a brief sketch of commutative ring theory which includes pertinent definitions along with main results without proof (but with ample references), and Part II, the 180 examples. The examples do cover a very large range of topics. Although most of them appear elsewhere, they are enhanced by a fairly complete listing of their properties. Example 67, for instance, is M. Hochster's counterexample to the polynomial cancellation problem, and it lists a number of properties of the two rings that were not given in the original paper Proc. Amer. Math. Soc. 34 (1972), no. 1, 81 - 82; MR 45 #3394. Some of the examples appear more than once, since many rings exhibit more than one interesting property. (R=Kx, y, z is used in Examples 6 and 22.) The examples are grouped into areas, but a drawback is that these have not been labeled and separated off. In addition, the Index is for Part I and definitions only, and this means that searching for a specific example with certain properties can be time consuming. The book can be used as a supplement to one of the standard texts in commutative ring theory, and it does appear to complement the monograph by I. Kaplansky Commutative rings, Allyn and Bacon, Boston, Mass., 1970; MR 40 #7234; second edition, Univ. Chicago Press, Chicago, Ill., 1974; MR 49 #10674. --Reviewed by Jon L. Johnson Thompson's group T is a finitely presented infinite simple group. Two non-zero commutative rings with unity, one a subring of the other, but with different unities. Let$R={\bf Z}/10{\bf Z}$,$S=2R$, then$R$and$S$are commutative rings with unity,$S$is a subring of$R$, but the identity element of$S$isn't the identity element of$R$. If we view$R$as$\lbrace0,1,\dots,9\rbrace$with operations modulo 10, so$S=\lbrace0,2,4,6,8\rbrace$, then the multiplicative identity in$S$is 6. This works more generally if$\gcd(m,n)=1$,$R={\bf Z}/mn{\bf Z}$, and$S=mR$. It works even more generally if$A$and$B$are non-zero commutative rings with unity,$R=A\times B$, and$S=A\times\lbrace0\rbrace$. A polynomial, solvable in radicals, whose splitting field is not a radical extension (of$\bf Q$). Let$f(x)$be any cyclic cubic, that is, any cubic with rational coefficients, irreducible over the rationals, with Galois group cyclic of order 3. Then$f(x)=0$is solvable in radicals (every cubic is), so the splitting field$K$of$f$over$\bf Q$is contained in a radical extension of$\bf Q$, but$K$is not itself a radical extension of$\bf Q$. The degree of$K$over$\bf Q$is 3, so for$K$to be radical over$\bf Q$it would have to be an extension of$\bf Q$by the cube root of some element of$\bf Q$, but such extensions are not normal. • So this just reminds us of the correct definition of a radical extension ... Jun 22 '10 at 10:55 While a finite abelian group is determined by its character table, this is not true for (finite) nonabelian groups. E.g., the dihedral and quaternion groups of order 8 (or more generally two nonabelian groups of order p3 for a prime p) are nonisomorphic but have the same character table. • R. Brauer asked whether knowing the character table, together with the "power maps" (knowing which conjugacy classes contain powers of elements frm any given conjugacy class) was enough to determine the group. E.C. Dade produced examples in the first issue of Journal of Algebra to show that Brauer's question still had a negative answer ( though Brauer's extra information distinguishes the different non-Abelian p-groups of order$p^{3}$. Jan 22 '15 at 3:59 If$f$and$g$are relatively prime in${\mathbf Q}[X]$then the mapping${\mathbf Q}[X]/(fg) \rightarrow {\mathbf Q}[X]/(f) \times {\mathbf Q}[X]/(g)$given by$h \bmod fg \mapsto (h \bmod f, h \bmod g)$is a ring isomorphism. This is a special case of the Chinese remainder theorem. If we replace${\mathbf Q}[X]$with its subring${\mathbf Z}[X]$, which is a UFD, then for relatively prime$f$and$g$in${\mathbf Z}[X]$the mapping${\mathbf Z}[X]/(fg) \rightarrow {\mathbf Z}[X]/(f) \times {\mathbf Z}[X]/(g)$given by$h \bmod fg \mapsto (h \bmod f, h \bmod g)$is a ring homomorphism, but it is not necessarily an isomorphism since it need not be surjective (though it is injective). For example, if$f = X-1$and$g = 1+X+\cdots + X^{n-1}$where$n > 1$then the natural mapping $${\mathbf Z}[X](X^n-1) \rightarrow {\mathbf Z}[X]/(X-1) \times {\mathbf Z}[X]/(1+X+\cdots + X^{n-1})$$ does not have$(0,1)$in its image. The reason is that if$f(X) \in {\mathbf Z}[X]$is mapped to$(0,1)$then$f(X) = (X-1)g(X)$for some$g(X) \in {\mathbf Z}[X]$and then$1 = (\zeta_p-1)g(\zeta_p)$for any prime$p$dividing$n$, which says$\zeta_p-1$is a unit in${\mathbf Z}[\zeta_p]$, and that's false. • But curiously enough,$\mathbb{Z}\left[X\right] / \left(\Phi_{n} \Phi_m\right) \to \mathbb{Z}\left[X\right] / \left(\Phi_n\right) \times \mathbb{Z}\left[X\right] / \left(\Phi_m\right)$is an isomorphism whenever$n/m$is not of the form$p^a$for some prime$p$and some$a \in \mathbb{Z}\$. (See, e.g., Theorem 1 in Dresden's Resultants of cyclotomic polynomials, home.wlu.edu/~dresdeng/papers/Res.pdf .) Oct 26 '14 at 16:11

This quasigroup is not isomorphic to any loop (i.e. quasigroup with identity):

* | a   b   c
-------------
a | a   c   b
b | c   b   a
c | b   a   c