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I am struggling with the following problem:

Find a group $G$ such that whenever $m, n, k \geq 2$ are natural numbers, then there exist $a, b \in G$ such that the order of $a$ is $m$, order of $b$ is $n$, and order of $ab$ is $k$.

The group $G$ necessarily has to be non-commutative and infinite.

Watson
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pizet
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  • You are right. I have corrected it. – pizet Feb 01 '16 at 10:30
  • Do you know how to do this if you only need such an $a$ and $b$ to exist for a single triple $(m,n,k)$? – Eric Wofsey Feb 01 '16 at 10:36
  • The obvious answer product of $G_{m,n,k}$ with $G_{m,n,k}$ given by the generators $a$, $b$ and relations $a^m = 1$, $b^n = 1$, $(ab)^k = 1$ is good enough? (with different generators $a$, $b$ for different $G_{m,n,k}$). – Martín-Blas Pérez Pinilla Feb 01 '16 at 10:46
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    You can have a look here. – Watson Feb 01 '16 at 10:53
  • The link you sent is not very helpful.Of course the product of $G_{m,n,k}$ with $G_{m,n,k}$ given by a presentation is an answer. I thought there is another way to describe it. – pizet Feb 01 '16 at 12:03
  • @pizet : the link I gave shows that if $q \equiv 1 \pmod {8mnk}$ is a prime power, then $A_{m,n,k} = \text{PSL}2(\Bbb F_q)$ has an element $a$ of order $m$, $b$ of order $n$ and $ab$ of order $k$. Then the product of all the $A{m,n,k}$ is what you want (if I'm not mistaken). – Watson Feb 01 '16 at 12:29
  • I do not understand why (according to the link) it suffices to construct elements elements $a, b \in \text{SL}(2,q)$ such that $a$, $b$, and $ab$ have orders $2m$, $2n$, and $2k$, respectively. – pizet Feb 01 '16 at 13:08
  • Related: http://math.stackexchange.com/questions/456764 – Watson Jun 23 '16 at 12:02

1 Answers1

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Suppose that for any integers $m,n,k>1$, we are able to find a group $A_{m,n,k}$ such that there exist $a \in A_{m,n,k}$ of order $m$, $b \in A_{m,n,k}$ of order $n$ with $ab \in A_{m,n,k}$ of order $k$. Then the group $$G = \prod\limits_{m,n,k>1} A_{m,n,k}$$ satisfies your request.

According to this answer, such groups $A_{m,n,k}$ exist. Indeed, you can consider a prime power $q = q(m,n,k)$ such that $q \equiv 1 \pmod{8mnk}$. (You can even take $q$ to be a prime number).

The answer in the link shows how to construct elements $x,y \in \text{SL}_2(\Bbb F_q)$ such that $x$ has order $2m$, $y$ has order $2n$, and $xy$ has order $2k$. Then $A_{m,n,k} = \text{PSL}_2(\Bbb F_{q})$ is the desired group.

Indeed, by definition, $$\text{PSL}_2(\Bbb F_{q}) = \text{SL}_2(\Bbb F_{q})/\text{SZ}_2(\Bbb F_{q}) = \text{SL}_2(\Bbb F_{q})/\{±I_2\}$$ since $q>2$.

If $π : \text{SL}_2(\Bbb F_{q}) \longrightarrow \text{PSL}_2(\Bbb F_{q})$ is the canonical homomorphism, then $a=π(x),b=π(y) \in A_{m,n,k}$ have order $m,n$ respectively and $ab$ has order $k$. This is because the only elements of order $≤2$ in $\text{SL}_2(\Bbb F_{q})$ are $I_2$ and $-I_2$. See below for a proof.

More precisely : we know that $x^m$ is of order $2$. Then $x^m=-I_2$ and $π(x^m)=a^m=1$ ; in particular the order of $a$ divides $m$. You can show that $a^k=1 \implies 2m \mid 2k \implies m ≤ k$. Therefore the order of $a$ is exactly $m$. Similar reasonings can be done for $b$ and $ab$.

Finally, a group satisfying your request is : $$ \color{purple}{\prod\limits_{m,n,k>1} \text{PSL}_2(\Bbb F_{q(m,n,k)}\;\;)} $$ with $q(m,n,k) \equiv 1 \pmod{8mnk}$ a prime power.

NB :

  1. In general, if you quotient a group $G_0$ by a normal subgroup of order $2$ (here $\{±I_2\}$), the image $π(g)$ of an element of order $2r$ doesn't have to be $r$. See for instance $g = (1,3) \in \Bbb Z/4\Bbb Z \times \Bbb Z/6\Bbb Z =: G_0$ has order $4$, whereas $π(g) = (1,0) \in G_0/(\{0\} \times 3\Bbb Z/6\Bbb Z) \cong \Bbb Z/4\Bbb Z \times \Bbb Z/3\Bbb Z$ has also order $4$.

  2. If $x \in G = \text{SL}_2(\Bbb F_{q})$ verifies $x^2=I_2$, then $x=I_2$ or $x = -I_2$. Instead of doing explicit calculations, we may look at the minimal polynomial $m_x(T)$ of $x$, which divides $T^2-1 = (T-1)(T+1)$. Therefore $m_x$ is split over $\Bbb F_q[T]$, and $x$ is diagonalizable, with $1$ or $-1$ as eigenvalues : you can write $x=SDS^{-1}$, with $S$ invertible, $D$ diagonal. Finally, $\det(x)=1$ yields $D=I_2$ or $D=-I_2$, i.e. $x=I_2$ or $x=-I_2$.

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Watson
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