By Dylan Wilson


2010-06-21 23:10:07 8 Comments

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)

30 comments

@Lior Bary-Soroker 2014-05-14 07:15:23

My favorite counter example in Galois theory is the field extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$.

Here are some cases to which it provides a counter example:

  1. It is a non-Galois extension (so providing counter example to "every extension is Galois").

  2. The Galois group of its Galois closure is non-abelian.

  3. Although the intersection of it with $\mathbb{Q}(\zeta_3 \sqrt[3]{2})$ is $\mathbb{Q}$, these fields are not linearly disjoint.

[Here $\zeta_3=e^{\frac{2\pi i}{3}}$.]

@QqQqqqq 2011-06-19 09:02:52

Regarding Schur's lemma:

For a finite group $G$ and $V$ a finite-dimensional irreducible representation of $G$ over a field $K$, there exist endomorphisms of this representation that are not scalar multiples of the identity. For example, take $G=\mathbb{Z}_4$, $K=\mathbb{R}$, and $\rho:\mathbb{Z_4}\rightarrow GL(\mathbb{R}^2)$ given by

$$\rho(1)=\begin{pmatrix} 0 & -1 \\ 1 & \ \ 0 \end{pmatrix}$$

Then since $\rho(1)$ has no real eigenvalues the representation is irreducible. But on the other hand, $\mathbb{Z}_4$ is abelian and $\rho(1): \mathbb{R}^2\rightarrow\mathbb{R}^2$ is an endomorphism of this representation.

This is why it is important $K$ be algebraically closed.

@Ryan Reich 2011-06-20 15:42:15

If $V$ is any representation of anything, and if $M$ is any matrix without eigenvalues in $K$ that commutes with everything, then it is a counterexample to Schur's lemma without algebraic closure (the proof of the lemma tells you this). In particular, if the group is abelian these are easy to find. What if the group is not abelian?

@Yiftach Barnea 2010-06-25 07:29:08

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?

@Mark Sapir 2010-10-09 22:07:36

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.

@Geoff Robinson 2014-05-14 19:28:08

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.

@Filippo Alberto Edoardo 2017-01-23 13:14:48

The Krull topology on an absolute Galois group is not the profinite topology.

For instance, let $G_\mathbb{Q}=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ for some fixed algebraic closure $\overline{\mathbb{Q}}/\mathbb{Q}$. It is easy to see that $G_\mathbb{Q}$ has uncountably many (normal) subgroups of index $2$, because they are $1-1$ with one-dimensional quotients of a $\mathbb{F}_2$-vector space of infinite dimension. Being of finite index, they are all open in the profinite topology, hence closed since a profinite group is compact when endowed with the profinite topology. Since there are only countably many quadratic extensions of $\mathbb{Q}$, one of the above subgroups cannot be closed in the Krull topology, by the fundamental theorem of infinite Galois theory.

@eltonjohn 2010-06-28 06:48:53

Please forgive me if someone has already posted this...

Let $X > Y > Z$ be a tower of groups with $Y$ and $Z$ being normal subgroups of $X$ and $Y$, respectively. $Z$ need not be a normal subgroup of $X$.

An example: $D_4 >$ Klein's $4$-group $> Z/2Z$.

@Dylan Wilson 2010-06-28 07:30:16

I just did this exercise in Dummit and Foote- nice! P.S. Big fan of Pinball Wizard and Tiny Dancer.

@mathcounterexamples.net 2014-12-14 18:16:38

A finitely generated soluble group isomorphic to a proper quotient group

Let $\mathbb{Q}_2$ be the ring of rational numbers of the form $m2^n$ with $m, n \in \mathbb{Z}$ and $N = U(3, \mathbb{Q}_2)$ the group of unitriangular matrixes of dimension $3$ over $\mathbb{Q}_2$. Let $t$ be the diagonal matrix with diagonal entries: $1, 2, 1$, put $H = \langle t, N \rangle$ and $w=\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$. Then the group $G=H/\langle w^2 \rangle$ is finitely generated soluble and isomorphic to a proper quotient subgroup.

For more details you can see here.

@Ken W. Smith 2011-06-20 13:56:24

Desmond MacHale wrote an article, "Minimum Counterexamples in Group Theory", Mathematics Magazine, 54 (1981), no. 1, 23–28; jstor. I've found this paper useful in an introductory algebra class and I like the philosophy of the paper, "Is X true? No, probably not. So what is a smallest counterexample?" A variation on the group theory (and Irish!) tune of MacHales appears here. A followup article is "Constructing a minimal counterexample in group theory" by Arnold Feldman, also in Mathematics Magazine (1985).

@Gerry Myerson 2010-06-22 06:14:20

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

@user100272 2016-10-31 04:19:54

I have been maintaining this small blog for a few months https://counterexamplesinalgebra.wordpress.com/

@Joshua Meyers 2015-10-22 17:59:50

A principal ideal with two non-associate generators (i.e. generators that are not unit multiples of each other).

In the ring $k[x,y,z]/(x-xyz)$, we have $([x])=([xy])$, but there is no unit $u$ such that $[xy]=u[x]$. See http://blog.jpolak.org/?p=534

@Geoff Robinson 2015-10-22 18:19:45

The (several) results of A.H. Schofield answering Artin's question (in the negative) by constructing skew-field extensions $K \subset L$ such that the right and left degrees are different deserve a mention.

@i707107 2015-07-31 21:34:25

I am also making this list for my record. The examples here are marvelous. I will put something that were not in this list. These examples are from Field Theory.

An Algebraic Extension of Infinite Degree.

$\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5, \cdots)$ over $\mathbb{Q}$.

A Nontrivial Finite Extension that is Isomorphic to the Ground Field.

Let $F=\mathbb{Q}(x)$ and $k=\mathbb{Q}(\sqrt x)$. Then $k$ is a degree-2 extension of $F$. However, they are isomorphic.

A Finite Extension which Contains Infinitely Many Subextensions.

Let $p$ be a prime. Let $F=\mathbb{GF}(p)(x,y)$ and $k=\mathbb{GF} (p) (x^{\frac 1 p},y^{\frac 1 p})$. For any $f(y)\in \mathbb{GF}(p)(y)$, $$ K=F(x^{\frac 1 p} f(y) + y^{\frac 1 p})$$ is a nontrivial subextension of $k$.

@Gerry Myerson 2015-10-23 06:10:53

The last is also a finite extension which is not a simple extension.

@Pace Nielsen 2015-01-22 04:00:26

Counter-example to the idea that algebraic duals cannot become simpler.

Consider the free $\mathbb{Z}$-module on a countable set $V=\mathbb{Z}^{(\omega)}$. The dual is $V^{\ast}\cong \mathbb{Z}^{\omega}$, a countable direct product of $\mathbb{Z}$'s, which is not free and also much bigger than $V$. Strangely, the double dual is $V^{\ast\ast}\cong V$!

@Michał Masny 2015-01-22 03:47:57

This is obscure, but Mogiljanskaja gave an example of two (and even an infinite sequence of) non-isomorphic semigroups such that their power semigroups are isomorphic. This resolved a question of Schein and Tamura. I don't remember the example now, but I should have the papers in my family house. If this is of any interest to anybody I could try to write it up since the papers are not readily available -- my copies came from Russia by snail mail and it cost a bit. There are two papers. The first shows an example of a sequence of semigroups and is quite simple. The second one adds one semigroup to the sequence and that's a bit more intricate, but still elementary. The semigroups can also easily be made commutative, which Mogiljanskaja remarks on.

@mathcounterexamples.net 2014-10-26 11:11:08

Examples of modules not having a basis.

It is well known that a vector space always has a basis. A module may not have a basis. Here are some examples:

  • The module $\mathbb{Z}/n\mathbb{Z}$ of the integers modulo $n$. This module has torsion.
  • The module $\mathbb{Q}$ of the rational numbers over the integers. This module is torsion-free.
  • The module $F[X]$ over the ring $F^\prime[X]$ of polynomials that have coefficient of $X$ equal to $0$. This module is finitely generated and torsion-free.

For more details, you can have a look here.

@truebaran 2014-10-26 15:59:18

This one is only about terminology, but while the topic is couterexamples in Algebra so it's tempting to give this one: Lie algebra is not an algebra.

@Tobias Kildetoft 2014-12-15 10:59:31

Sure it is. It is just not an associative algebra.

@YCor 2016-10-31 04:51:02

It's an algebra and it's not an algebra. Mathematicians seem to find practical to use "algebra" in several distinct meanings, which are more or less obvious to guess according to the context, when they are not specified. In practice, it is very common that people deal with together both associative unital algebras and Lie algebras, and call the first simply "algebras" and the second "Lie algebras" (e.g., search stuff about universal enveloping algebras, since it relates the two).

@Omar Antolín-Camarena 2014-10-26 14:48:29

I like Amnon Yekutieli's example of a module whose completion is not complete.

Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ an $A$-module. Algebraically you can define the completion $\hat{M}$ of $M$ as the inverse limit of the modules $M / I^k M$ (with the canonical quotient maps $M / I^{k+1} M \to M / I^k M$). There is a canonical module morphism $M \to \hat{M}$ and you can call $M$ ($I$-adically) complete if this is an isomorphism.

I used to think that the completion of an arbitrary module is complete! Rest assured that this is true if $A$ is Noetherian. But it does fail for the simplest example of a non-Noetherian ring: take $A = k[x_1, x_2, \ldots]$ the ring of polynomials in countably many variables, and $M = A$. For the ideal $I = \langle x_1, x_2, \ldots \rangle$ generated by the all variables, the completion $\hat{M}$ is, as one would expect from the finite dimensional case, the ring of power series in countably many variables (these power series should have only finitely many monomials of any given degree, so something like $\sum_i x_i$ does not count). However this module is not $I$-adically complete: indeed, look at the sequence of polynomials $\sum_{i=1}^n x_i^i$. If it did converge to a power series, by comparing coefficients, it is clear that the limit would have to be $\sum_{i=1}^\infty x_i^i$. (Since all power series in $I^k \hat{M}$ have only monomials of degree at least $k$, elements of the completion of $\hat{M}$ have a well-defined coefficient for any monomial.) But it does not in fact converge to that since it is easy to check that the tails, $\sum_{i=j}^\infty x_i^i$ do not lie in any $I^k \hat{M}$, i.e, you can't have an equality of the form $\sum_{i=j}^\infty x_i^i = m_1 g_1 + \cdots + m_l g_l$, where the $m_i$ are finitely many monomials: every term on the RHS mentions one of the finitely many variables present in the $m_i$, but there is no such "finite cover by variables" for the LHS.

I learned this example from Amnon Yekutieli's paper On Flatness and Completion for Infinitely Generated Modules over Noetherian Rings.

@Pavel Čoupek 2014-10-26 13:52:34

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), ...

@KConrad 2014-10-18 16:03:56

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.

@darij grinberg 2014-10-26 16:11:13

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 .)

@bof 2014-05-15 07:46:45

Here's one from universal algebra: the class of mono-unary algebras $\mathfrak A=(A,f)$ such that $f$ is a permutation with a unique fixed point does not have the unique factorization property for direct decomposition.

(Spoiler alert: if you hover over the gray box, an example of what can go wrong is revealed.)

For $n\equiv1\pmod3$ let $\mathfrak A_n=(A,f)$ where $|A|=n$ and $f$ is a permutation of order $3$ with a unique fixed point; then $\mathfrak A_{100}\cong\mathfrak A_{10}\times\mathfrak A_{10}\cong\mathfrak A_4\times\mathfrak A_{25}$, and $\mathfrak A_4,\mathfrak A_{10},\mathfrak A_{25}$ are indecomposable.

@Stahl 2014-10-18 16:32:17

If $K/\Bbb Q$ is a number field and $\mathcal{O}_K$ its ring of integers, $\mathcal{O}_K$ need not have a power basis as a $\Bbb Z$-module; i.e., $\mathcal{O}_K\neq\Bbb Z[\alpha]$ for any $\alpha\in\mathcal{O}_K$! An example is given by $K = \Bbb Q(\alpha)$, where $\alpha$ is a root of $T^3 - T^2 - 2T - 8$ (a $\Bbb Z$-basis is instead given by $\{1,\alpha,(\alpha^2 + \alpha)/2\}$).

See Keith Conrad's notes for more detail on this example.

@bof 2014-05-15 07:25:34

An example showing that, in the standard definition of a ring (without $1$), it won't do to replace the left and right distributive laws with the single law$$(u+v)\cdot(x+y)=u\cdot x+u\cdot y+v\cdot x+v\cdot y$$as was done on p. 18 of my old copy (July 1957 printing) of Kelley's General Topology.

(Spoiler alert: if you hover over the gray box, an example of what can go wrong is revealed.)

Take an additive group of order $3$, choose a nonzero element $c$, and define $x\cdot y=c$.

@darij grinberg 2014-10-26 16:01:37

Do you also mean to exclude the $0a = a0 = 0$ axiom?

@bof 2014-10-26 20:40:08

@darijgrinberg Kelley's General Topology, first printing, p. 18: "A ring is a triple $(R,+,\cdot)$ such that $(R,+)$ is an abelian group and $\cdot$ is a function on $R\times R$ to $R$ such that: the operation is associative, and the distributive law $(u+v)\cdot(x+y)=u\cdot x+u\cdot y+v\cdot x+v\cdot y$ holds for all members $x$, $y$, $u$, and $v$ of $R$."

@bof 2014-10-26 20:43:51

@darijgrinberg $0a=a0=0$ is not usually included among the axioms for a ring, since it is a consequence of the usual distributive laws together with the additive group properties.

@bof 2014-10-26 20:58:23

@darijgrinberg Of course you only have to add the axiom $0\cdot0=0$ to Kelley's axioms to get a correct definition of a ring.

@darij grinberg 2014-10-26 21:31:51

OK, I see what you mean. Well, I am used to including $0a=a0=0$ with the axioms, but I guess this is atypical.

@sms1 2014-05-15 03:44:13

Assertion: If $S$ is an associative ring with identity and $M, N$ are unital left $S$-modules, then $\hom_S(M,N)$ is a unital left $S$-module.

Th is is false in general. Consider the matrix ring $S:=M_2({\mathbb Z})$, and let $M={\mathbb Z}^2=N$ be equipped with natural $S$-module structure. We have $\hom_S(M,N)\hookrightarrow S$, as additive groups; moreover, if $\phi\in \hom_S(M,N)$, then $\phi(A{\bf v})=A(\phi({\bf v}))$ for every $A\in M_2({\mathbb Z})$ and ${\bf v}\in M$. Thus, the scalar matrices, and only those, are the elements in $\hom_S(M,N)$, since the only matrices that commute with all four of the elementary matrices are precisely the scalar matrices. By the isomorphism ${\mathbb Z}\cong {\mathbb Z} I_2$, we have $X:=\hom_S(M,N)\cong {\mathbb Z}$, and this makes $\hom_{\mathbb Z}(X,X)\cong {\mathbb Z}$.

Now, to make $X=\hom_S(M,N)$ into a unital left $S$-module, we need to give a unital ring homomorphism $S\rightarrow \hom_{\mathbb Z}(X,X)\cong {\mathbb Z}$. But there is no such. To see this, we notice that the elementary matrices in $S$ are nilpotent elements, and therefore, must be mapped to the unique nilpotent element $0\in {\mathbb Z}$. Since a ring homomorphism is additive, the only homomorphism $S\rightarrow {\mathbb Z}$ must be trivial.

@crskhr 2011-06-18 08:25:12

  • 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 $G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

@Per Alexandersson 2015-10-22 21:15:24

Math.se is coincidentally an actual Swedish math website also... math.se

@Geoff Robinson 2014-05-14 19:35:42

The quaternion group of order $8$ has a real irreducible character of degree $2,$ but the associated representation can not be realized over the real field.

@Sinan Sertoz 2011-07-06 11:43:31

Radical of a primary ideal is prime but not every ideal whose radical is prime is primary. Here is a cute counterexample: Let $I=(x^2,xy)\in F[x,y]$ where $F$ is a field. The radical $\sqrt{I}$ of $I$ is $(x)$ which is prime but $I$ is not primary; $xy\in I$, $x\not\in I$ but no power of $y$ belongs to $I$.

This is from page 154 of Commutative Algebra Vol. 1 by Zariski and Samuel. Now that I check, this is the 1975 printing which I bought on 1979. How time flies when you are having fun! :-)

@Abel Stolz 2010-08-11 13:40:39

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.

@Kimball 2010-08-11 13:13:47

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.

@Geoff Robinson 2015-01-22 03:59:12

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 $p^{3}$.

@Gerry Myerson 2010-08-11 06:01:51

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 $S=A\times\lbrace0\rbrace$.

@Gerry Myerson 2010-06-22 03:33:34

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.

@The Mathemagician 2010-06-22 05:46:41

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.

@Pete L. Clark 2010-06-27 23:16:08

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.

@Pete L. Clark 2010-06-27 23:16:49

(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...)

@Geoff Robinson 2014-05-17 20:28:57

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.

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