# Is the free product $\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}$ linear over $\mathbb{Z}$?

Let $H:=\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}=\langle p,q| q^n=1\rangle.$ I want to know if $H$ is a ($\mathbb{Z}$)linear group that is to say is there an injective homomorphism $f: H\to GL_m(\mathbb{Z})$ for $m\geq n.$ I asked the question on Math Stack Exchange (https://math.stackexchange.com/questions/1430677/is-the-free-product-mathbbz-mathbbz-n-mathbbz-a-linear-group) and Derek Holt suggested me to also ask it here. By advance thank you.

• Can't you use that it has a free subgroup of finite index? This subgroup has a faithful Z-linear representation and you can induce it to the whole group. Or am I missing something? Sep 12, 2015 at 10:39
• @BenjaminSteinberg Yes that sounds good! The normal closure of $\langle p \rangle$ is a free subgroup of rank $n$ and of index $n$ in $H$. Since ${\rm GL}_2({\mathbb Z})$ contains free subgroups, we get a faithful representation of $H$ of degree $2n$. I should have thought of that myself - but I was trying to find a representation of degree $n$. Sep 12, 2015 at 10:55
• I might as well make this an answer. Sep 12, 2015 at 11:51
• where can I find a detailed proof of the fact that a free group has a faithful representation over $\mathbb{Z}$ of degree $2?$ Sep 12, 2015 at 22:03
• In a free product any element of finite order is conjugate to an element of a free factor. So the kernel of the projection to Z/N has no element of finite order so is free. Sep 13, 2015 at 0:39

The group $Z*Z/n$ is virtually free: the kernel $K$ of the projection to $Z/n$ is free of index $n$ (say by the Kurosh theorem and since each finite order element is conjugate to an element of $Z/n$). Since a free group has a faithful representation over $Z$ of degree 2, the induced representation of this representation gives a faithful representation over $Z$ of your group of degree $2n$.