Problems from the Kourovka Notebook that undergraduate students can fully appreciate

Solution 1:

Problem 8.10(a) from the 8th edition (1982):

Is the group $G = \langle a, b \mid a^n=1, ab = b^3 a^3 \rangle$ finite or infinite for $n = 7$? All other cases known. See Archive, 7.7 and 8.10 b. (D. L. Johnson)

Remark:

• for $n=3$ the group has the order 6 (should be an easy exercise for a student to check this by hand and show that it's cyclic)

• for $n=6$ it has the order 9072 (perhaps not so easy to check this by hand, but can be done using computer).

• for $n=7$, the computer calculation runs too long without an answer.

• It is known that $G$ is infinite for:

  • $n = 15$ in [D. J. Seal, Proc. Roy. Soc. Edinburgh (A), 92 (1982), 181–192]
  • $n = 9$ (and $15$) in [M. I. Prishchepov, Commun. Algebra, 23 (1995), 5095–5117].

An example in GAP illustrates the problem:

gap> F:=FreeGroup("a","b");
<free group on the generators [ a, b ]>
gap> G:=F/ParseRelators(GeneratorsOfGroup(F),"a^3=1,ab=b^3*a^3");
<fp group on the generators [ a, b ]>
gap> Size(G); # could be easily done by hand
6
gap> G:=F/ParseRelators(GeneratorsOfGroup(F),"a^6=1,ab=b^3*a^3");
<fp group on the generators [ a, b ]>
gap> Size(G);
9072
gap> G:=F/ParseRelators(GeneratorsOfGroup(F),"a^7=1,ab=b^3*a^3");
<fp group on the generators [ a, b ]>
gap> IsFinite(G);
#I  Coset table calculation failed -- trying with bigger table limit
#I  Coset table calculation failed -- trying with bigger table limit
... GAP was interrupted ...

The message about the coset table calculation hitting the limit is often a slight hint towards the fact that it may be infinite, but that's far from being the evidence - it is still possible that the calculation will succeed after increasing the limit several times.

Thus, the problem for $n=7$ is still open...

Update: the answer to this question is given now in the 7th revision of the 18th edition of the Kourovka Notebook (http://arxiv.org/abs/1401.0300):

This group is infinite, because it contains the Fibonacci group $F(3, 7)$ as an index $7$ subgroup. This follows from Theorem 3.0 of (C. P. Chalk, Commun. Algebra 26, no. 5 (1998), 1511–1546) by standard technique for working with Fibonacci groups (G. Williams, Letter of 6 October 2015).

Solution 2:

Problem 17.76 from the 17th edition (2010):

Does there exist a finite group $G$, with $|G| > 2$, such that there is exactly one element in $G$ which is not a commutator? (D. MacHale)

Solution 3:

Problem 15.99 from the 15th edition (2002):

Let $f(n)$ be the number of isomorphism classes of finite groups of order $n$. Is it true that the equation $f(n) = k$ has a solution for any positive integer $k$? The answer is affirmative for all $k \le 1000$ [G. M.Wei, Southeast Asian Bull. Math., 22, no. 1 (1998), 93–102]. (W. J. Shi)

Solution 4:

Problem 18.49 from the 18th edition (2014):

Let $n \in \mathbb{N}$. Is it true that for any $a, b, c \in \mathbb{N}$ satisfying $1 < a, b, c \le n-2$ the symmetric group $S_n$ has elements of order $a$ and $b$ whose product has order $c$? (S. Kohl)

Update: the 7th revision of the 18th edition of the Kourovka Notebook (http://arxiv.org/abs/1401.0300) says that the answer is positive and refers to the preprint A note on the product of two permutations of prescribed orders by Joachim König. From the abstract:

We prove a conjecture by Stefan Kohl on the existence of triples of permutations of bounded degree with prescribed orders and product 1. This result leads to an existence result for covers of the complex projective line with bounded degree and prescribed ramification indices.

See also https://mathoverflow.net/q/118092/

Update 2: 19th edition of the Kourovka notebook (https://arxiv.org/abs/1401.0300) gives a reference: A note on the product of two permutations of prescribed orders, J.König, Eur. J. Comb., 57 (2016), 50–56, https://doi.org/10.1016/j.ejc.2016.03.006.