Draw the Cayley graph of the group \(\mathbb{Z} * \mathbb{Z}_2 = \langle a, b \mid b^2 \rangle\).
Solution
Let \[ G=\mathbb Z*\mathbb Z_2 =\langle a,b\mid b^2=1\rangle. \] Use the generating set \[ \{a,a^{-1},b\}. \]
1. Every vertex of the Cayley graph has three incident edge ends: an \(a\)-edge leaving it, an \(a\)-edge entering it, and one \(b\)-edge.
From a vertex \(g\) there are edges to \[ ga,\qquad ga^{-1},\qquad gb. \] Since \(b=b^{-1}\), the \(b\)-edge from \(g\) to \(gb\) is the same undirected edge when viewed from \(gb\).
2. The \(a\)-edges form disjoint bi-infinite lines, one for each right coset of \(\langle a\rangle\).
Starting from \(g\) and traversing only \(a\)-edges gives precisely \[ \{ga^n:n\in\mathbb Z\}, \] with consecutive vertices joined by \(a\)-edges. No nonzero power of \(a\) is trivial, so this is a bi-infinite line.
3. The \(b\)-edges join these \(a\)-lines in a perfect matching, and the resulting graph has no cycles.
Every vertex lies on exactly one \(b\)-edge because \(b^2=1\). If there were a cycle, reading its labels and cancelling immediate reversals would give a nonempty reduced word in the free product \[ \mathbb Z*\mathbb Z_2 \] representing the identity, contradicting the free-product normal-form theorem.
4. Thus the Cayley graph is the infinite \(3\)-regular tree.
By 1 every vertex has valence three, and by 3 the connected Cayley graph is acyclic. Equivalently, draw a bi-infinite \(a\)-line through the identity; from every vertex attach one \(b\)-edge to a new bi-infinite \(a\)-line, and repeat indefinitely. This is the requested drawing up to graph isomorphism.