Results from abstract algebra which look wrong (but are true)

Question

There are many statements in abstract algebra, often asked by beginners, which are just too good to be true. For example, if $N$ is a normal subgroup of a group $G$, is $G/N$ isomorphic to a subgroup of $G$? As an experienced mathematician, we see immediately that there is no reason for this to be true — even without thinking about this in detail. Often we can quickly come up with counterexamples. Sometimes, it is hard to find counterexamples.

Many questions fall into this category, for example:

• If $f : R \to S$ is a ring homomorphism and $I \subseteq R $ is an ideal, is then $f(I) \subseteq S$ an ideal? (SE/2200335)
• $\DeclareMathOperator\Aut{Aut}$If $G,H$ are groups, do we have $\Aut(G \times H) \cong \Aut(G) \times \Aut(H)$? (SE/1236571)
• Is every submodule of a finitely generated module also finitely generated? (SE/83078)
• If $A$ is an abelian group with $A^3 \cong A$, does this imply $A^2 \cong A$? (MO/10128)
• If $A$ is an abelian group with $A \oplus \mathbb{Z}^2 \cong A$, does this imply $A \oplus \mathbb{Z} \cong A$? (MO/218113)
• If $G$, $H$ are groups whose group algebras $ \mathbb{Q}[G]$, $\mathbb{Q}[H]$ are isomorphic, are then $G$, $H$ isomorphic? (SE/1342851)
• see also MO/23478 for common false beliefs in mathematics

But my question is actually about situations where, for some strange reason, our first gut feeling is not correct and a wrong-looking statement turns out to be true. Examples will be abundant, which is why I want to restrict this question to examples coming from abstract algebra (you are welcome to open similar questions for other branches and flavors of mathematics, and please let me know if there are already questions of this type).

Here are some examples which come to my mind:

• Every group homomorphism $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}$ is a finite linear combination of projections. In fact, $(\mathbb{Z}^{ \mathbb{N}})^* \cong \mathbb{Z}^{\oplus \mathbb{N}}$. (Specker 1950)
• If $A$, $B$ are finitely generated abelian groups (more generally, finitely generated modules over a commutative Noetherian ring) and $f : A \to A \oplus B$, $g : A \oplus B \to B$ are homomorphisms such that $0 \to A \xrightarrow{f} A \oplus B \xrightarrow{g} B \to 0$ is exact, then it is split exact.
• If $A$, $B$, $C$ are finite groups such that $A \times B \cong A \times C$, then $B \cong C$. (SE/3579745)
• every negation of the examples mentioned above, for example: There is an abelian group $A$ with $A \cong A^3$ and $A \not\cong A^2$. (However, I am more interested in "positive" results.)

I am looking for statements in abstract algebra where this is your reaction when you learn that they are actually true.

Please try to include a reference for the statement and proof.

Answer 1

The free group with infinitely many generators is a subgroup of the free group with two generators.

Answer 2

Let $G$ be a finite group and $n \mid |G|$.

If $S = \{x \in G : x^n = 1\}$ contains exactly $n$ elements, then $S$ is a subgroup of $G$.

There seems no a priori reason to expect $S$ to be a subgroup if $G$ is non-abelian. The proof (see Iiyori and Yamaki - On a conjecture of Frobenius (MSN)) uses the classification of finite simple groups.

Answer 3

Every finite simple group can be generated by at most $2$ elements. This is another famous consequence of the classification.

Answer 4

Every element of a finite simple non-abelian group is a commutator. This is the positive solution to the Ore conjecture (see Liebeck, O’Brien, Shalev, and Tiep - The Ore conjecture) and uses the classification.

Answer 5

Every finite index subgroup of a finitely generated profinite group is open. The converse is obvious, but this direction was quite surprising to me. This is a result of Nikolov and Segal and uses the classification of finite simple groups.

Answer 6

The Nielsen-Schreier theorem that subgroups of free groups are free might have seemed surprising from am algebraic view given the analogue for many other algebraic structures is false. While this is easy to prove topologically, the original algebraic proof is in my view just an algebraic translation of the topological proof that for some strange reason, preceded the topological one.

Answer 7

As fields, the algebraic closures of the fields ${\bf Q}_p$ are isomorphic, and are isomorphic to the complex numbers.

Answer 8

For a group with finitely many elements of finite order, the set of elements of finite order is a subgroup.

Answer 9

The projective dimension of ${\mathbb C}(x,y,z)$ as a module over ${\mathbb C}[x,y,z]$ is two if the continuum hypothesis holds, and three otherwise.

Answer 10

In combinatorial group theory, loosely speaking almost any problem one can imagine, in full generality, turns out to be undecidable. This includes the word problem, the isomorphism problem, the triviality problem, etc. Here's an example of a very general problem which nevertheless is decidable.

In the mid 1960s and early 1970s, the "equation problem" or "Diophantine problem" for free semigroups (and groups) was studied. Given a fixed free group $F$ (for simplicity, say of rank $2$ on the generators $a$ and $b$), this asks: let $w_1(a, b, X,Y,Z,\dots)$ and $w_2(a,b, X,Y,Z,\dots)$ be two words written in the generators $a,b$ and their inverses, together with some "variables" $X, Y, Z, \dots$ and their formal inverses. Are there words $w_X, w_Y, \dots, w_Z \in F$ such that the equation $$ w_1(a, b, w_X, w_Y, w_Z, \dots) = w_2(a,b, w_X, w_Y, w_Z, \dots) $$ holds true in $F$? For example, is there a solution to $XbXY^{-1} = Zb$ in the free group on $a$ and $b$? (Yes, e.g. take $X = Y = Z = a$). The same problem can be asked for free semigroups instead (in which case one just omits the inverses -- these are just called "word equations").

The "real" Diophantine problem, i.e. the problem of determining whether a polynomial over $\mathbf{Z}$ have integer roots, or Hilbert's Tenth Problem, was proved undecidable in general by Matiyasevich in 1970. For a long time it was believed that Hilbert's Tenth Problem should be reducible to the Diophantine problem in free (semi)groups, and thereby prove this latter problem undecidable too.

But this was not to be. In 1977, Makanin proved that free semigroups have decidable Diophantine problem, and a few years later, he also proved that free groups have decidable Diophantine problem. His solutions are incredibly intricate, but have been generalised to all hyperbolic groups by Dahmani & Guirardel (this latter solution is based on Razborov diagrams; Razborov was, like Makanin, also a student of Adian's, and worked to make a geometric version of Makanin's combinatorial arguments).

This is one of few instances of a very general problem which is decidable in combinatorial group theory. The first time I saw it, I was certain I had misread "undecidable" as "decidable", given the general gloom of undecidability in this area (cf. the Adian-Rabin theorem). I think this qualifies!

Answer 11

There exists a finitely-generated infinite group with only two conjugacy classes, a difficult result of Osin.

Answer 12

In a finite Frobenius group, the set of all fixed point free elements together with the identity forms a subgroup.

This might not have such a shocking effect to us, since usually when we first hear about Frobenius groups, it's precisely with the goal of proving this statement, but then again, usually in group theory, when some subsets in a large class of groups don't have a "natural" reason for being subgroups, sooner or later they won't be (and also, the above becomes false in infinite Frobenius groups).

Answer 13

It seems to me appropriate to name the following totally unexpected result, which is too good to be true, yet is true.

Are there only finitely many finite groups with $m$ generators of exponent $n$, up to isomorphism (Restricted Burnside problem)?

In the case of the prime exponent $p$, this problem was extensively studied by A. I. Kostrikin during the 1950s, prior to the negative solution of the general Burnside problem. The case of arbitrary exponent has been completely settled in the affirmative by Efim Zelmanov, who was awarded the Fields Medal in 1994 for his work.

Answer 14

Let $F$ be a non-abelian free group and let $G=\prod_\omega F$ be the direct product of infinitely many copies of $F$. Then the abelianisation of $G$ has torsion (of order $2$), by a theorem of Kharlampovich and Myasnikov ["Implicit function theorem over free groups and genus problem", Knots, braids, and mapping class groups—papers dedicated to Joan S. Birman].

Granted, this is an example of an unreasonably bad phenomenon, not an unreasonably good phenomenon, but I still couldn't believe my ears when I was told it (by Lars Louder).

Answer 15

The Auslander–Buchsbaum theorem that every regular local ring is a unique factorization domain.

I should say that the first time I saw this theorem stated, I was not immediately surprised, but that was because I did not yet have enough experience with commutative algebra to have a well-developed intuition either way. Somehow, the Auslander–Buchsbaum theorem became more amazing to me the more I learned.

Answer 16

I suppose there is a case for saying that Jordan's theorem on finite complex linear groups might be such a result: there is a function $f: \mathbb{N} \to \mathbb{N}$ such that for every $n \in \mathbb{N}$, every finite subgroup $G$ of ${\rm GL}(n,\mathbb{C})$ has an Abelian normal subgroup $A$ with $[G:A] \leq f(n).$ This is well known to fail if we try to replace $\mathbb{C}$ by an algebraically closed field of characteristic $p > 0$.

Answer 17

Quite a few things in the Hopf algebra world are surprising:

• Takeuchi's theorem: Every connected graded bialgebra is a Hopf algebra. (No finiteness assumptions!) Takeuchi was actually more general: If $H$ is an $\mathbb N$-graded bialgebra and its $0$-th graded component $H_0$ is Hopf, then $H$ is Hopf.

• The Cartier-Milnor-Moore theorem: Every cocommutative graded (= $\mathbb N$-graded) bialgebra in characteristic $0$ is isomorphic to the universal enveloping algebra of its primitive space (= space of primitive elements). This means that understanding cocommutative graded bialgebras in characteristic $0$ is essentially equivalent to understanding graded Lie algebras. Commutative graded bialgebras can be understood likewise by duality. You can replace "graded" by "filtered", and the theorem still holds with appropriate changes. To see proper wild behavior, you need to go to the neither-commutative-nor-cocommutative case, or to positive characteristics (or work over rings). I consider the proof of Cartier-Milnor-Moore (by studying Eulerian idempotents) to be full of surprises as well, but maybe more combinatorial than algebraic ones.

• The Nichols-Zoeller theorem: If $H$ is a finite-dimensional Hopf algebra over a field, and $A$ is a Hopf subalgebra of $H$, then $H$ is a free left $A$-module and a free right $A$-module. (Perhaps this is somewhat less surprising if you think of it as generalizing Lagrange's theorem for finite groups, but there are so many more Hopf algebras than groups!) Generalized even further by Skryabin (2006).

• Zelevinsky's theory of PSH algebras, showing that every $\mathbb Z$-Hopf algebra that satisfies certain positivity properties is isomorphic to a tensor product of degree-stretched copies of the ring of symmetric functions.

Once you get to symmetric functions, the surprises start multiplying: I find the "Schur polynomial = alternant divided by Vandermonde determinant" identity surprising no matter how many different proofs I see; the Littlewood-Richardson rule in its many forms; the semistandard tableaux forming a section of the plactic monoids while also indexing a basis of irreducible $\operatorname{GL}_n$-modules; ... But maybe surprises are somewhat less surprising when they come from combinatorics, as we are used to think of algebra as formal manipulations and of combinatorics as a jungle full of life?

And then, back in algebra, there is of course the Fundamental Theorem of Galois theory.

Answer 18

In ZFC, the complex number field has $2^{2^{\aleph_0}}$ automorphisms, whereas the real number field has just one, the identity.

Answer 19

1. There is some theory about maximal valuation rings (a special type of ring with linearly ordered ideals, not necessarily a domain) and then there are almost-maximal valuation rings which is of course a relaxation. But if you look for an almost-maximal valuation ring which is not maximal, it has to be a domain for some reason (Gill - Almost Maximal Valuation Rings).

2. The descending chain condition on right ideals of a ring implies the ascending chain condition on right ideals. The descending chain condition on principal right ideals does not imply the ascending chain condition on principal right ideals: instead it implies the ascending chain condition on principal left ideals (Jonah - Rings with the minimum condition for principal right ideals have the maximum condition for principal left ideals).

Answer 20

If $k$ is an algebraic number field then for every positive integer $n$ there exist infinitely many field extensions of $k$ of degree $n$ having no proper subfields over $k$.

Answer 21

A subring of a Noetherian ring need not be Noetherian.

Given all the stability properties that Noetherian rings enjoy, this may sound surprising at first, but it becomes much more obvious if you think about the fact that any domain can be embedded in a Noetherian ring - its field of fractions.

Perhaps more surprising though is that this can also fail for finitely generated rings (or finitely generated algebras over a field): the subring $\mathbb Z[2X,2X^2,2X^3,\dots]$ of $\mathbb Z[X]$ is not Noetherian, and neither is the subring $k[XY,XY^2,XY^3,\dots]$ of $k[X,Y]$ for any field $k$.

Answer 22

A theorem of Bass: For a ring $R$, every left $R$-module has a projective cover if and only if $R$ satisfies the descending chain condition on principal right ideals.

Answer 23

An example might be that the category of abelian groups is hereditary. That is, every complex of abelian groups is quasi-isomorphic to the (graded) direct sum of its cohomologies. Although this is very straightforward, I know of at least one person (me) who was surprised by this statement when learning homological algebra

Answer 24

That there exist finitely presentable non-Hopfian groups. [I still remember my shock when I was first learned this result!]

A group G is Hopfian if every surjective homomorphism $\phi:G\to G$ is in fact bijective. Most common-or-garden finitely generated groups are Hopfian, e.g. linear groups, finitely generated residually finite groups, and hyperbolic groups are all Hopfian. For non-finitely generated groups we can mimic Hilbert's hotel, so for example the infinite product $\mathbb{Z}\times\mathbb{Z}\cdots$ is non-Hopfian.

The Baumslag-Solitar group $\operatorname{BS}(2, 3)=\langle a, b\mid b^{-1}a^2b=a^3\rangle$ is non-Hopfian, with the relevant map being $a\mapsto a^2, b\mapsto b$.

Answer 25

In the same vein of the statement the OP included:$$G\times H \cong G\times K \Longrightarrow H\cong K$$ for product of finite groups, which can be rephrased as "product in finite groups is cancellative", a similar property holds true for product "powers".

$$ \underbrace{G\times \cdots \times G}_{n\text{ times} } \cong \underbrace{H\times \cdots \times H}_{n\text{ times}} \Longrightarrow G\cong H$$

In other words to take powers of finite groups is cancellative.

Answer 26

A simple module $S$ over a finite dimensional algebra $A$ over an algebraically closed field $K$ such that there exists a non-split short exact sequence $0\rightarrow S \rightarrow X \rightarrow S \rightarrow 0$ has infinite projective dimension. This is the strong no loops conjecture and it is not known whether the assumption on the field can be removed, see for example Igusa, Liu, and Paquette - A proof of the strong no loop conjecture.

Answer 27

Let $G$ be a profinite group and let $C_G$ the category of discrete $G$-modules. Define the cohomological dimension of $G$ as

$$\mathrm{cd}(G)=\inf\;\{n \geq 0 \mid \mathrm{H}^q(G,A)=0\; \text{for every $q>n$ and torsion module $A \in C_G$} \}.$$ Clearly if finite, this number is less or equal to the strict cohomological dimension

$$\mathrm{scd}(G)=\inf\;\{n \geq 0 \mid \mathrm{H}^q(G,A)=0\; \text{for every $q>n$ and $A \in C_G$} \}.$$

Suprisingly we have the upper bound $$\mathrm{scd}(G)\leq \mathrm{cd}(G)+1.$$

Answer 28

We say that a binary operation $*$ on a set $X$ is left-distributive if it satisfies the identity $x*(y*z)=(x*y)*(x*z).$ The free left-distributive algebra generated by a countably infinite set embeds into the free left-distributive algebra generated by a 2 elements.