TITLE: Is the free product $\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}$ linear over $\mathbb{Z}$?
QUESTION [9 upvotes]: 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.

REPLY [14 votes]: 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 $2n$.