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I'm interested in unsolved problems in group theory, which are easy enough to state to a beginner. This question is very much in the same spirit as this similar question about calculus.

A problem should:

  • Only use the language of a typical first course in group theory. This is somewhat subjective but generally we want to be talking only about subgroups, normal subgroups, generating sets, Lagrange's theorem, Cauchy's theorem, Cayley's theorem, Sylow theorems, etc.

  • Be an unsolved problem.

Here is an example (determining the infinitude of the free Burnside groups). Define the group $B(n,m)$ as the quotient group of the free group on $n$ generators by the normal subgroup of all $m$-th powers. When is $B(n,m)$ finite? When is it infinite? In particular, we do not know if $B(2,5)$ is finite or infinite.

What are some other examples?

Robin
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    Just to point out that you could easily state the Burnside Problem without reference to free groups, which are not always covered in first courses in group theory: for which $m$ and $n$ is it true that all groups $G$ that can be generated by $m$ elements and satisfy $g^n=1$ for all $g \in G$ are finite. – Derek Holt May 19 '25 at 16:17
  • At first I thought this was about graph theory, and I was very confused because I didn't recognize any of the stuff you were talking about as a beginner in graph theory. – nick012000 May 20 '25 at 07:30

13 Answers13

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Have a look at the Kourovka Notebook, it is available on Arxiv: https://arxiv.org/abs/1401.0300

It is about unsolved problems and deciding which ones are appropriate for beginners is up to discussion.

Vicfred
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Beginners in group theory are usually familiar with the fact that every group of order $n$ lives as a copy in $S_n$ (Cayley's theorem). An unsolved problem is whether every group of order $n!$ has a subgroup of order $n$.

Robin
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Kan't
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The following is a conjecture of Thompson:

Let $G$ be a finite non-abelian simple group. Then there exists a conjugacy class $C$ of $G$ such that every $g \in G$ can be written in the form $g =xy$ for some $x,y \in C$.

I note that the Ore conjecture (which is known to be true) is a corollary, so Thompson's conjecture seems to be a very difficult problem. For some recent progress, see for example here.

testaccount
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  • Huh. Given that we have a classification of all finite simple groups, that this is open is really surprising. – JoshuaZ May 20 '25 at 16:33
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    @JoshuaZ: Basically, just because we know what the simple groups are, does not mean it is straightforward to understand their structure. There are still many open questions about finite simple groups which are difficult to answer. Here is another open problem: for which finite non-abelian simple groups $G = \langle x, y \rangle$ with $|x| = 2$ and $|y| = 3$? https://doi.org/10.1017/S1446788724000016 – testaccount May 21 '25 at 01:11
13

You didn't mention abelian groups or direct sums but those are typically mentioned in a first course of group theory. The following problem is mentioned in Fuchs, Infinite Abelian Groups II and attributed to A. L. S. Corner:

Does there exist an abelian group such that every nonzero direct summand is a direct sum of infinitely many nonzero subgroups?

Unfortunately I haven't found a better reference for this problem (and it's possible it has been solved since then) so if anyone knows more about it please edit this answer.

Carla_
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9

I quote from https://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/102104#102104

I think you could give an accessible K-12 formulation of the definition of a group (as a group of permutations, for instance) and of an integral group ring. The Zero Divisor Conjecture (Kaplansky, 1940) then states, in one version, that if $G$ is a torsion-free group then the group ring $\mathbb{Z}[G]$ has no zero divisors besides the number $0$.

Gerry Myerson
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6

The Wielandt-Kegel theorem states that all finite dinilpotent groups $G=AB$, i.e., where $A,B$ are nilpotent subgroups of $G$, are solvable. Although this is true for many infinite groups, e.g., for finitely-generated linear groups, or groups $G=AB$ with $A$ and $B$ abelian subgroups, this is open in general.

Conjecture: Every dinilpotent group is solvable.

Dietrich Burde
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One problem (Problem 4.65.) from the Kourovka notebook that I like a lot, even though it is not really group theory but number theory:

$\frac{p^q-1}{p-1}$ never divides $\frac{q^p-1}{q-1}$ for distinct primes $p,q$. This would simplify the proof of the odd order theorem.

Robin
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schiepy
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    It is an interesting problem, but unfortunately a solution would not really simplify the proof of the odd order theorem in a significant way. See the answer here: https://mathoverflow.net/questions/281043/feit-thompson-conjecture – testaccount May 19 '25 at 14:23
  • Ah, thanks for the info. I just looked at the notebook and found this conjecture with a note saying it would simplify the proof. – schiepy May 19 '25 at 18:40
4

Let $f(n)$ be the number of groups of order $n$ up to isomorphism.

Is

$$f(ab) \geq f(a)f(b)$$

for all $a, b$?

It’s true for $a,b$ coprime, and it’s also true for $ab \leq 2048$. Whether it’s always true is still unsolved.

Robin
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3

I like Thompson conjecture, a generalization of Ore's theorem, which have been stated by @testaccount.

Other things about simple groups are not known. For example, not every sporadic group has a simple construction. For example, some are automorphisms of exceptional structures, such as the Conway group $Co_3$ which is the automorphisms group of the Leech lattice. However we lack of constructions for some groups, suggesting there might exist some undiscovered exceptional strctures. For example, every construction of the Monster are very complicated and "not canonical" (like people did though constructions that matched the properties the Monster group should have). I believe it is the same for the Baby Monster.

Similarly, the Moonshine conjecture was proved by Richard Borcherds by making the Monster act on some vertex algebra. Borcherds is asking whether this so called algebra could have a simpler construction, see Problem 1 of this paper.

Enguerrand Moulinier
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3

Is it true that an arbitrary finitely presented group has either polynomial or exponential growth? The Grigorchuk group is finitely generated and has subexponential non-polynomial growth but it does not have a finite presentation. For an introduction to group growth rates with references, this one place to start.

Robin
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JoshuaZ
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    I do not think the notion of group presentations is normally discussed in undergraduate classes. However, one can state an open problem about growth which does not require discussing presentations (the gap conjecture): If a group $G$ has growth strictly slower than $\exp(\sqrt{n})$, then it has polynomial growth. – Moishe Kohan May 21 '25 at 01:50
  • @MoisheKohan I first learned about group presentations as an undergrad, and when I taught abstract algebra we did discuss it, but admittedly only briefly. I don't know how common it is; I'd agree certainly that it isn't universally taught. Your point about the gap conjecture is a good one, and is in that context probably worth its own separate answer. – JoshuaZ May 21 '25 at 01:58
  • I also learned group presentations as an undergrad, while studying algebraic topology. But this is uncommon. – Moishe Kohan May 21 '25 at 02:21
3

For which finite non-abelian simple groups $G$ is the following true?

$G = \langle x,y \rangle$ for some $x,y \in G$ with $|x| = 2$ and $|y| = 3$.

This problem is still open in some cases where $G$ is an orthogonal group $\operatorname{P \Omega}_n(q)$ with $n$ even.

See for example the following paper: "M.A. Pellegrini, M.C. Tamburini Bellani, The (2,3)-generation of the finite simple odd-dimensional orthogonal groups" https://doi.org/10.1017/S1446788724000016

testaccount
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How many groups are there of order $n$? (See e.g., MSE this post by Qiaochu Yuan for some discussion.)

In particular, how many groups are there of order 2048?

Kimball
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3

Higman's polynomials on residue classes ("PORC") conjecture: let $f(n)$ be the number of groups of order $n$.

Then for any natural number $m$ and all sufficiently large primes $p$, can $f(p^m)$ always be given as a family of polynomials in $p$, depending on what $p$ is modulo some other number?

It's true for $m \leq 7$:

$$\begin{align} f(p)&=1 \\ f(p^2)&=2 \\ f(p^3)&=5 \\ f(p^4)&=15 && \text{except $p=2$} \\ f(p^5)&=2p + 61 + 2 \gcd (p-1, 3) + \gcd(p-1, 4)&& \text{except $p=2,3$} \\ f(p^6)&=3 p^2+39 p+344+24 \operatorname{gcd}(p-1,3) && \text{except $p=2,3$} \\ & \quad +11 \operatorname{gcd}(p-1,4)+2 \operatorname{gcd}(p-1,5) \\ f(p^7)&= 3 p^5+12 p^4+44 p^3+170 p^2+707 p+2455 && \text{except $p=2,3,5$} \\ & \quad +\left(4 p^2+44 p+291\right) \operatorname{gcd}(p-1,3) \\ & \quad +\left(p^2+19 p+135\right) \operatorname{gcd}(p-1,4) \\ & \quad +(3 p+31) \operatorname{gcd}(p-1,5) +4 \operatorname{gcd}(p-1,7) \\ & \quad +5 \operatorname{gcd}(p-1,8)+\operatorname{gcd}(p-1,9) \\ \end{align}$$

Does this continue?

Robin
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