= Solution
For the finite state space $S$, the generator or <Q-matrix> $G=(g_{ij})$ is defined by
$$
g_{ij}=\lim_{h\downarrow0}
\frac{\mathbb P_i(X_h=j)-\mathbf1_{\{i=j\}}}{h}.
$$
Its off-diagonal entries are nonnegative jump rates and $g_{ii}=-\sum_{j\ne i}g_{ij}$. The <transition semigroup of a continuous-time Markov chain> is $P(t)=e^{tG}$.
An <invariant distribution of a continuous-time Markov chain> is a probability row vector $\pi$ satisfying $\pi P(t)=\pi$ for every $t\geq0$. Differentiation at zero gives $\pi G=0$; conversely, $\pi G=0$ implies $\pi e^{tG}=\pi$. Thus
$$
\boxed{\pi\text{ is invariant}\Longleftrightarrow\pi G=0.}
$$
Invariant distributions need not be unique: each <closed communicating class> supports one, and their convex combinations are invariant. A finite irreducible chain has a unique invariant distribution.
Let $0=T_0<T_1<\cdots$ be the arrival times of the independent rate-$\lambda$ <Poisson process>. The increments $T_{n+1}-T_n$ are independent exponential variables of rate $\lambda$. Conditional on $Y_n=X_{T_n}=i$, the <Markov property>, time homogeneity, and independence of the next increment give
$$
\begin{aligned}
\mathbb P(Y_{n+1}=j\mid Y_0,\ldots,Y_n)
&=\int_0^\infty\lambda e^{-\lambda t}P_{ij}(t)\,dt\\
&=(P_\lambda)_{ij}.
\end{aligned}
$$
Hence $(Y_n)$ is a discrete-time Markov chain. Its transition matrix is the <Poisson-sampled continuous-time Markov chain> resolvent
$$
\boxed{P_\lambda
=\int_0^\infty\lambda e^{-\lambda t}e^{tG}\,dt
=\lambda(\lambda I-G)^{-1}.}
$$
If $\pi G=0$, then $\pi e^{tG}=\pi$ and integration gives $\pi P_\lambda=\pi$. Conversely, if a probability row vector $\nu$ satisfies $\nu P_\lambda=\nu$, multiplying
$$
\nu\lambda(\lambda I-G)^{-1}=\nu
$$
on the right by $\lambda I-G$ yields $\nu G=0$. Therefore the sampled chain and the original chain have \b[exactly the same invariant distributions].
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