Suppose we have integer matrices $A_1,\ldots,A_n\in\operatorname{GL}(n,\mathbb Z)$. Define $\varphi:F_n\to\operatorname{GL}(n,\mathbb Z)$ by $x_i\mapsto A_i$. Is there an algorithm to decide whether or not $\varphi$ is injective?

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    $\begingroup$ The corresponding problem for semigroups is undecidable for $n\geq 3$. I don't know what is known for groups. It is probably open and undecidable. $\endgroup$ Commented Sep 1, 2011 at 0:50
  • $\begingroup$ Benjamin - I'd be interested in a reference for the semi-group case, if you have one. $\endgroup$
    – HJRW
    Commented Sep 1, 2011 at 6:01
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    $\begingroup$ This was proved by Klarner, Birget and Satterfield in IJAC in 1991 for something like n=5 and improved to 3 and upper triangular in iml.univ-mrs.fr/~cassaign/publis/freeness.ps.gz $\endgroup$ Commented Sep 1, 2011 at 13:52

3 Answers 3


For $n=1, 2$ the answer is "yes" since the group is virtually free, for $n\ge 3$ the answer is not known (an open problem).

Edit. In fact even for two $n\times n$-matrices the problem is open. Moreover the solution of the following ``easier" problem is not known: for which algebraic integers $\lambda$ the matrices $\left(\begin{array}{ll} 1 & 2\\\ 0 & 1 \end{array}\right)$ and $\left(\begin{array}{ll} 1 & 0\\\ \lambda & 1 \end{array}\right)$ generate a free group (see this paper, for example). The fact that this problem is easier follows from the trivial observation that the group generated by these two matrices is isomorphic to some effectively computable group of $n\times n$-integer matrices for some $n\ge 2$ (depending on the degree of the algebraic number $\lambda$).

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    $\begingroup$ It's not clear for me that's it's easier. It might happen that for each n there's an algorithm in GL(n,Z) to detect free groups (as in the initial question) but that this algorithm does not depend recursively on n. $\endgroup$
    – YCor
    Commented Sep 3, 2011 at 21:43
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    $\begingroup$ @Yves: The size of the matrices is the degree of the $\lambda$. There is currently no hope to resolve the problem even if the degree is 3. $\endgroup$
    – user6976
    Commented Sep 4, 2011 at 4:48

Here are some general facts that may be relevant.

Given a finitely presented group $G$ and a representation $\rho:G\to GL_n(\mathbb{Z})$, there is no algorithm which is uniform in $n$ that decides whether or not $\rho$ is injective.

However, this leaves open the possibility that there is such an algorithm for particular $n$. (It's easy for $n=2$, when the group is virtually free. I believe nothing is known for $n>2$.) Also, the examples we construct are not free groups, so it may be possible to say something in that case.

In another direction, given a finite presentation for a group $G$ and a solution to the word problem in $G$, one can algorithmically determine whether or not $G$ is a free group.


Some comments on the question 'Is it decidable whether or not a collection of integer matrices generates a free group?'

Given a finite set of matrices S over a field, the problem of testing whether the group H generated by S contains a free non-abelian subgroup is decidable. An algorithm solving the problem as well as its implementation (in Magma) available. Notice that the algorithms does not construct a free non-abelian subgroup in H but justifies its existence. As to testing freeness of finitely generated linear groups then the problem has quite a long history. I may recommend as a start point the paper by John Dixon Can.J Math, v. 37, n. 2, 1985, 238-259 (see p. 240 there, and then follow the references).

  • $\begingroup$ I cannot find this paper! Do you have a link? $\endgroup$ Commented Jul 26, 2021 at 11:07

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