Theorem 4.17: if conjugation on a free abelian normal subgroup has an eigenvalue off the unit circle then the group has a free subsemigroup of rank two

Let $G$ be a group with a normal subgroup $A$ carrying an isomorphism $A \cong \mathbb{Z}^r$ — so $A$ is free abelian of rank $r$. Let $g \in G$ and let $T$ be an $r \times r$ integer matrix representing conjugation by $g$ through that isomorphism: writing $\iota$ for the isomorphism $\mathbb{Z}^r \to A \le G$, for every $z \in \mathbb{Z}^r$

If the complexification of $T$ has an eigenvalue $\varphi$ — a nonzero $v \in \mathbb{C}^r$ with $Tv = \varphi v$ — whose modulus is not $1$, then $G$ contains a free subsemigroup of rank two.

Reading the pieces. $\mathbb{Z}^r$ appears as Multiplicative (Fin r → ℤ), the additive group written multiplicatively, so the isomorphism hypothesis says exactly "free abelian of rank $r$"; ofAdd and toAdd are relabelings, not exponentials. The matrix acts on column vectors from the left, $(Tz)_i = \sum_j T_{ij} z_j$. The hypothesis on $\varphi$ is a disequality $\lVert\varphi\rVert \neq 1$, not $\lVert\varphi\rVert > 1$; Rosenblatt reduces the case $\lVert\varphi\rVert < 1$ to the other by replacing $g$ with $g^{-1}$. The conclusion is the published Chou.HasFreeSubsemigroupOfRankTwo: some pair of elements makes distinct positive words take distinct values. No letter stands for an inverse, so this is a free subsemigroup claim, not a free subgroup claim.

The conjugation hypothesis, precisely. It constrains only the direction $x \mapsto g x g^{-1}$. Since the isomorphism is onto $A$, it says exactly that conjugation by $g$ maps $A$ into $A$ and is given by $z \mapsto Tz$ in coordinates. At $z = 0$ it holds automatically.

Normality is load-bearing, and the statement is vacuous for small $r$. Together with the conjugation hypothesis, normality forces $z \mapsto Tz$ to be onto $\mathbb{Z}^r$, hence $\det T = \pm 1$. Two consequences follow, both easy to miss. Normality cannot be dropped: without it the remaining hypotheses are satisfiable at $r = 1$, with $T = (2)$ and $\varphi = 2$. And with it the hypotheses are contradictory for $r \le 1$ — at $r = 0$ there is no nonzero $v$, and at $r = 1$ invertibility forces $T = (\pm 1)$, so $\varphi = \pm 1$ and $\lVert\varphi\rVert = 1$. The theorem therefore has content only from $r = 2$ on.

The hypotheses are satisfiable there: take $G = \mathbb{Z}^2 \rtimes \mathbb{Z}$ with $T = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}$, whose eigenvalue $(3+\sqrt 5)/2$ has modulus greater than $1$. The hypotheses also force $g \notin A$ and $G$ nonabelian.