I have been reading about random walks on Cayley graphs of groups lately and stumbled across the walk on the modular group $\mathbb{Z}/(2\mathbb{Z}) * \mathbb{Z}/(3\mathbb{Z})$, where $*$ denotes the free product of two groups.
A lot of papers say that this is trivially a transient walk, still I do not understand why that is. I am looking for a somewhat elementary proof of the statement, i.e. starting from the definition $G(x,x) < \infty$, where $G$ is Green's function defined by $$G(x,y|z) := \sum_{n \geq 0} p^{(n)}(x,y)z^n \quad \text{with} \quad G(x,y) := G(x,y|1).$$
Any help would be appreciated!
Let $e$ be the identity in the group $\Gamma:=\mathbb{Z}/(2\mathbb{Z}) * \mathbb{Z}/(3\mathbb{Z})$.
The solution below does a bit more than requested, as we will prove a bound (5) for $G(x,e)$ that is sharp.
Every element $x \ne e$ in $\Gamma$ can be represented as a word $$x=a^{m_0}b^{d_1}a\,b^{d_2} a\, b^{d_3}a \cdots b^{d_k}a^{m_1} \,, \tag{1}$$ where $a$ generates $\mathbb{Z}/(2\mathbb{Z})$ and $b$ generates $\mathbb{Z}/(2\mathbb{Z})$. Here the exponents $m_0,m_1$ are in $\{0,1\}$ and the exponents $d_i$ are in $\{1,2\}$. (The case $x=a$ corresponds to $k=0$.) For such $x$, if $k \ge 1$ we let $B(x)=k$ denote the number of $b$ powers that appear in $x$, so $B(ab^2)=B(aba)=1$. Let $A(x)=k+m_0+m_1-1$ denote the number of $a$'s that appear in $x$. Also, let $A(e)=B(e)=0$ and $A(a)=1, B(a)=0$. Every $x \in \Gamma$ has three neighbors in the Cayley graph, one obtained by erasing or adding $a$, and two obtained by erasing (or adding) $b,b^2$.
Write $G_n(x)=G_n(x,e)=\sum_{k=0}^n p^{(k)}(x,e)$. We will prove inductively a bound of the form $$\forall n \ge 0, \quad G_n(x) \le Cr^{A(x)}s^{B(x)} \,, \tag{2}$$ where $r,s \in (0,1)$ and $C>1$ will be determined later.
Proof of (2): $\,$ Clearly (2) holds for $n=0$, so suppose that $n \ge 1$ and (2) holds with $n-1$ in place of $n$. For every $x \ne e$ in $\Gamma$, we have $$G_n(x)=\frac13 \sum_{j=1}^3 G_{n-1}(y_j) \,,$$ where $y_j$ are the neighbors of $x$ in the Cayley graph. (Note that for $x=e$ we must add 1 to the right hand side.) If $x$ ends with an $a$ (so $m_1=1$ in (1) or $x=a$), then the induction hypothesis gives $$G_n(x) \le Cr^A(x)s^B(x)\cdot\frac{r^{-1}+2s}{3} \,. \tag{3}$$ If $x$ ends with $b^{\gamma_k}$ (so $k \ge 1$ and $m_1=0$ in (1)), then the induction hypothesis gives $$G_n(x) \le Cr^A(x)s^B(x)\cdot\frac{s^{-1}+1+r}{3} \,. \tag{4}$$ Choose $r,s \in (0,1)$ so that $$r^{-1}+2s=3=s^{-1}+1+r, \; \,\text{namely,} \;\, r=2/3 \;\, \text{and} \; s=3/4 \, \,. $$ Then (3) and (4) verify the induction step for $x \ne e$. For $x=e$, the induction hypothesis gives $$G_n(e) \le 1+C(r+2s)/3 =1+\frac{13}{18}C \,,$$ so picking $C=\frac{18}{5}$ yields $G_n(e) \le C$ and completes the induction step. Thus (2) is proved. $\hspace{6.5in} \Box$
Taking the limit as $ n \to \infty$ in (2) yields $$\forall x \in \Gamma, \quad G(x,e) \le \frac{18}{5} \cdot (2/3)^{A(x)} (3/4)^{B(x)} \,. \tag{5}$$
$ $ Addendum: $\,$ This is not needed for the original question, but (5) is actually an equality. indeed, if the RHS of (5) is denoted by $\psi(x)$, then the difference $h(x):=\psi(x)-G(x)$ is a non-negative harmonic function on $\Gamma$, i.e. $$\forall x \in \Gamma, \quad h(x)=\frac13 \sum_{j=1}^3 h(y_j) \,,$$ where $y_j$ are the neighbors of $x$ in the Cayley graph. Since $h(x)$ tends to zero as $\text{dist}(x,e) \to \infty$, the maximum principle implies that $h \equiv 0$.
Remark If one knows about electrical networks (see, e.g., Chapter 2 in [1] or Chapter 9 in [2]) then one can infer that $G(x,x)=18/5$ directly from the formula $G(x,x)=\text{deg}(x) \cdot R_{\text{eff}}(x,\infty)$ and the parallel-series laws that imply the effective resistance $R_{\text{eff}}(x,\infty)$ equals $6/5$.
[1] Lyons, Russell, and Yuval Peres. Probability on trees and networks. Vol. 42. Cambridge University Press, 2017. https://www.yuval-peres-books.com/probability-on-trees-and-networks/
[2] Levin, David A., and Yuval Peres. Markov chains and mixing times. Vol. 107. American Mathematical Soc., 2017., https://www.yuval-peres-books.com/markov-chains-and-mixing-times/