TITLE: Is every $C_0$ semigroup on a Hilbert space automatically a $C_0$ group on a larger space?
QUESTION [7 upvotes]: Let $\{T(t),t\ge 0\}$ be a $C_0$ semigroup on a Hilbert space $X$, does that exist a larger Hilbert space $Y$ such that $X\subset Y$, and $T(t)$ extend to a  $C_0$ group $T'(t)$ (so $t<0$ make sense now) on $Y$?  
Edit: Here "extend" means that $(T(t)x, y)=(T'(t)x, y)$ for all $x, y\in X$, which is weaker than its usual meaning. If the semigroup $T(t)$ is a contraction, then the conclusion is true, see Theorem 8.1 (p.29) in "Harmonic analysis of operators on Hilbert spaces" written by  Sz. Nagy, C. Foias,  H. Bercovici and L. Kérchy.

REPLY [5 votes]: An operator $A$ is a group generator if and only if both $A$ and $-A$ are semigroup generators. So under your assumptions, if we denote by $\tilde{A}$ the generator of the extension $\{\tilde{T}(t),t\ge 0\}$ of $\{T(t),t\ge 0\}$ to $Y$, then $-\tilde{A}$ would be a semigroup generator, too.
If by "extend" you mean the usual thing (viz, that $X$ is left invariant under $\{\tilde{T}(t),t\ge 0\}$ and the restriction to $X$ of $\{\tilde{T}(t),t\ge 0\}$ is $\{T(t),t\ge 0\}$ again), then $X$ would be left invariant under $\{T(t),t\le 0\}$, too, and the generator of this semigroup would necessarily be the part in $X$ of $-\tilde{A}$. So, the analytic semigroup $\{T(t),t\ge 0\}$ would actually imbed into a group $\{T(t),t\in \mathbb R\}$ on $X$ as well.
Now, an analytic semigroup on a Banach space $X$ can imbed into a group if and only if $X$ is finite dimensional its generator is bounded.