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Suppose that I have a group $G$ with a finite number of generators $(a_1,a_2,\ldots,a_n)$, and the orders $|a_1|,|a_2|,\ldots,|a_n|$ are all finite. Does it follow that the order of the group $|G|$ is necessarily finite?

I'm not sure how to even approach proving/disproving this problem. To try disproving it, I tried finding an infinite group that had a finite number of generators each with finite order, but everything I tried didn't seem to work:

  • $\mathbb{Z}$ has generators of infinite order
  • $GL_n(\mathbb{R})$ the set of invertible $n\times n$ matrices, also has generators of infinite order
  • The infinite cyclic group has a finite number of generators, but of infinite order

I don't really know many examples of groups (infinite or finite), so I got bummered here.

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  • $\begingroup$ You may be interested in looking up Dietzmanns's Lemma: "In any group a normal finite subset consisting of elements of finite order generates a finite group". books.google.co.uk/… $\endgroup$ – Evgeny T Dec 24 '16 at 11:37
  • $\begingroup$ Note that 'the' infinite cyclic group is isomorphic to $\Bbb{Z}$. Have you also considered subgroups of $\operatorname{GL}_n(\Bbb{R})$ generated by matrices of finite order? $\endgroup$ – Servaes Dec 24 '16 at 11:49
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Define $f:\mathbb{Z}\to\mathbb{Z}$ by $$f(n)=\begin{cases} n+1&\text{ if $n$ is even} \\ n-1&\text{ if $n$ is odd}\end{cases}$$ and $g:\mathbb{Z}\to\mathbb{Z}$ by $$g(n)=\begin{cases} n-1&\text{ if $n$ is even} \\ n+1&\text{ if $n$ is odd.}\end{cases}$$

Then $f$ and $g$ are both permutations of $\mathbb{Z}$, and have order $2$ in as elements of the group of all permutations of $\mathbb{Z}$. But notice that $$(g\circ f)(n)=\begin{cases} n+2&\text{ if $n$ is even} \\ n-2&\text{ if $n$ is odd}\end{cases}$$ so $g\circ f$ is a permutation of infinite order (as you iterate it on even and odd integers, you just keep adding or subtracting $2$, respectively). It follows that the group of permutations generated by $f$ and $g$ is infinite, even though $f$ and $g$ both have finite order.

(In fact, it turns out that this is the same as silvascientist's example. The explicit representation as permutations makes it clear that all the words you can form by alternating $g$ and $f$ really are distinct elements, since you can just compute what they are as functions.)

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Consider the free group $G$ on $a$ and $b$, mod the relation that $a^2 = b^2 = e$. Then $G$ admits sequences of arbitrary length, namely $a$, $ab$, $aba$, $abab$, $ababa$, $\ldots$.

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    $\begingroup$ I actually thought of this group as a possible counterexample, but I'm completely new to group theory, so I'm not sure how to prove that all the sequences are distinct? $\endgroup$ – Joshua Lin Dec 24 '16 at 4:07
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    $\begingroup$ @JoshuaLin This is isomorphic to the group of maps $f:\mathbb Z\rightarrow \mathbb Z$ of the form $f(x)=a\pm x$ for integer $a$. Yould take $a$ is the map $x\mapsto -x$ and $b$ is the map $x\mapsto 1-x$. This makes it easier to prove that the elements are distinct. $\endgroup$ – Milo Brandt Dec 24 '16 at 4:11
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    $\begingroup$ It's also the infinite dihedral group. $\endgroup$ – pjs36 Dec 24 '16 at 4:14
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    $\begingroup$ @TannerSwett Eric Wofsey's example provides a more concrete representation, with $g$ and $f$ generating the group and having the order indicated. As he mentions, "the explicit representation as permutations makes it clear that all the words you can form by alternating $g$ and $f$ really are distinct elements, since you can just compute what they are as functions". At any rate, if in any group we can have two elements of order two which generate it, and alternating sequences of these are distinct, then this certainly holds in the free group on these two elements, modding that they have order 2. $\endgroup$ – Sir Jective Dec 24 '16 at 5:57
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    $\begingroup$ An different alternative representation could be to let $a$ and $b$ be reflections in two lines in the plane whose mutual angle is an irrational multiple of $360^\circ$. That is, for example, in matrices, $a=({}^1\,{}_{-1})$ and $b=({}^{0.6}_{0.8}\,{}^{\;0.8}_{-0.6})$ in $GL(\mathbb Q,2)$. $\endgroup$ – Henning Makholm Dec 24 '16 at 11:45
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In plane Euclidean geometry, the group generated two reflections in two parallel lines provides an easy example. The product of the two reflections is a nonzero translation, which has infinite order.

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    $\begingroup$ nice example with geometry. $\endgroup$ – mesel Dec 24 '16 at 15:21
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I. The dihedral group $D_n$ is a group of order $2n$ generated by two elements $a_n,b_n$ of order $2.$ Then $a=(a_1,a_2,a_3,\dots)$ and $b=(b_1,b_2,b_3,\dots)$ are elements of order $2$ in the direct product $\prod_{n=1}^\infty D_n,$ and they generate a subgroup $D=\langle a,b\rangle$ which is infinite because each $D_n$ is a homomorphic image of $D$ via the projection map.

II. Any permutation can be expressed as a product of two involutions [W. R. Scott, Group Theory, Prentice–Hall, 1964, Exercise 10.1.17]. In particular, given a permutation $\pi$ of infinite order, we can write $\pi=\alpha\beta$ for some involutions $\alpha$ and $\beta,$ and then $G=\langle\alpha,\beta\rangle$ is an infinite group generated by two elements of order $2.$ For a concrete example where $\pi$ is an infinite cycle in $\operatorname{Sym}(\mathbb N),$ take $$\alpha=(1\ 2)(3\ 4)(5\ 6)(7\ 8)\cdots,$$ $$\beta=(1)(2\ 3)(4\ 5)(6\ 7)\cdots,$$ $$\pi=\alpha\beta=(\cdot\ \cdot\ \cdot\ 7\ 5\ 3\ 1\ 2\ 4\ 6\ 8\ \cdot\ \cdot\ \cdot).$$

III. Any countable group can be embedded in a $2$-generator group with generators of prescribed orders $p\ge3$ and $q\ge2$ [F. Levin, Factor groups of the modular group, J. London Math. Soc. 43 (1968), 195–203]. Hence, any countable group can be embedded in a group generated by three elements of order $2.$

IV. See Burnside's problem for examples of finitely generated infinite groups in which each element has finite order.

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