Let $X_n, n\geq1$ be a Markov chain with state space $\{1,\dots,6\}$ and transition matrix $$ P = \left[ \begin{array} \\ 0 & 0 & 1-\alpha & \alpha & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1-\alpha & 0 & \alpha & 0 \\ \end{array} \right] $$
Find $\rho_{xy}=P(T_y<\infty| X_0=x)$ and $G(x,y)=\mathbb{E}[N(y)|X_0=x]$ such that $T_y$ is the first passage time and $N(y)$ is the number of visits to state $y$.
Clearly states $\{1,3,4\}$ are recurrent and states $\{2,5,6\}$ are transient. So for the first set $G(x,y)=\infty$
We know that the following holds $$ \begin{align} \rho_{xy} &= P(T_y<\infty| X_0=x) \\ &= \sum_{n=1}^{\infty}P(T_y=n| X_0=x) \\ \\ G(x,y) &=\mathbb{E}[N(y)|X_0=x]\\ &=\mathbb{E}[\sum_{n=1}^{\infty}I\{X_n=y\}|X_0=x] \\ &=\sum_{n=1}^{\infty}P^n(x,y) \end{align} $$ such that $P^n$ is the n-step ahead transition function. Taking for example $x=5,y=5$, $$ \begin{align} \rho_{55} &= \sum_{n=1}^{\infty}P(T_5=n| X_0=5) \\ G(5,5) &= \sum_{n=1}^{\infty}P^n(5,5) \end{align} $$ We may notice that we have only one way back from $5$ into $5$, $5\to 6\to 5$ and thus we have zero probabilities for reaching 5 in even number of steps thus
$$G(5,5) = \sum_{k=1}^{\infty}P^{2k}(5,5) = \sum_{k=1}^{\infty}\alpha^k = \frac{1}{1-\alpha}$$
However, if I find $\rho_{xy}$ through the system of equations, $$ \rho_{ij}=P(i,j)+\sum_{z\neq j}P(i,z)\rho_{z,j} $$ and apply the theorem that says that for transient states $G(x,y)=\rho_{xy}/(1-\rho_{yy})$, I get $\rho_{55}=\rho_{66}=\rho_{65}=\alpha$, $\rho_{56}=1$ and $G(5,5)=\alpha/(1-\alpha)$ as the result.
What am i doing wrong?
I would appreciate any insights.