= Solution
A <strongly continuous semigroup> on the complex <Banach space> $E$ is a family $T_t\in\mathcal L(E)$ with $T_0=I$, $T_{s+t}=T_sT_t$ and $\|T_tf-f\|\to0$ as $t\downarrow0$ for every $f\in E$. Its <semigroup generator> is the <linear operator>
$$
D(Z)=\left\{f\in E:\lim_{h\downarrow0}\frac{T_hf-f}{h}\text{ exists in }E\right\},\qquad
Zf=\lim_{h\downarrow0}\frac{T_hf-f}{h}.
$$
For $f\in D(Z)$, boundedness of $T_t$ and the semigroup identity give
$$
\frac{T_hT_tf-T_tf}{h}=T_t\frac{T_hf-f}{h}\longrightarrow T_tZf.
$$
Thus <semigroup commutation with its unbounded generator> means
$$
\boxed{T_tD(Z)\subseteq D(Z),\qquad ZT_tf=T_tZf\quad(f\in D(Z)).}
$$
There is no assertion that $Z$ is defined on every vector. The orbit is differentiable on this domain, with $\frac d{dt}T_tf=T_tZf$.
A <contraction semigroup> additionally satisfies $\|T_t\|\leq1$ for every $t\geq0$. For $\operatorname{Re}\lambda>0$, define the <Bochner integral>
$$
L_\lambda f=\int_0^\infty e^{-\lambda t}T_tf\,dt.
$$
<Strong continuity> gives a continuous integrand, and its <norm> is bounded by $e^{-t\operatorname{Re}\lambda}\|f\|$, so the integral exists and
$$
\|L_\lambda\|\leq\frac1{\operatorname{Re}\lambda}.
$$
Changing variables in $T_hL_\lambda f$ yields
$$
T_hL_\lambda f=e^{\lambda h}
\left(L_\lambda f-\int_0^h e^{-\lambda s}T_sf\,ds\right).
$$
After subtracting $L_\lambda f$ and dividing by $h$, the limit is $\lambda L_\lambda f-f$. Consequently
$$
\boxed{L_\lambda(E)\subseteq D(Z),\qquad
(\lambda I-Z)L_\lambda f=f.}
$$
For $f\in D(Z)$, integrate the <derivative> of $e^{-\lambda t}T_tf$ over $[0,\infty)$. Its limit at infinity is zero by the contraction bound, and its value at zero is $f$. Therefore
$$
L_\lambda(\lambda I-Z)f
=-\int_0^\infty\frac d{dt}(e^{-\lambda t}T_tf)\,dt=f.
$$
Thus this is also the <Laplace-transform formula for a semigroup resolvent>, with $L_\lambda=(\lambda I-Z)^{-1}$.
To prove that $Z$ is a <closed operator>, suppose $f_n\in D(Z)$, $f_n\to f$ and $Zf_n\to g$. Fix one $\lambda$ with positive real part. The identity just proved and boundedness of $L_\lambda$ give
$$
f=\lim_nL_\lambda(\lambda f_n-Zf_n)=L_\lambda(\lambda f-g).
$$
Its range lies in $D(Z)$, and applying $\lambda I-Z$ shows $Zf=g$. Thus the graph is closed. Linearity follows directly from linearity of the defining difference quotient, so \b[$Z$ is a closed <linear operator>].
The closedness conclusion also holds without the contraction assumption. For a general <strongly continuous semigroup>, the <Uniform boundedness principle> gives a uniform operator bound on each compact time interval. Integrating the generator-orbit identity gives $T_tf_n-f_n=\int_0^tT_sZf_n\,ds$. Under $f_n\to f$ and $Zf_n\to g$, pass to the limit to obtain $T_tf-f=\int_0^tT_sg\,ds$. Divide by $t$ and let $t\downarrow0$; <strong continuity> makes the right side converge to $g$, so $f\in D(Z)$ and $Zf=g$.
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