 Solution

Direct multiplication gives $T^2=P^2=I$, $TP=PT=-I$, andThus conjugation by each generator of $G^{\prime }$ preserves $G$; so does conjugation by every product of the generators. Therefore $\boxed{G⊴G^{\prime }}$.

Using $TB(x)=B(-x)T$ and its analogue for $P$, every group word can be reduced to $B(x)S$ with $S\in \{I,T,P,TP\}$. Consequently the quotient has at most four cosets. They are distinct: a nonidentity diagonal matrix among $T,P,-I$ cannot equal $B(x)$, since a diagonal $B(x)$ requires $\sinh x=0$ and hence $B(x)=I$. ThereforeThe three nonidentity cosets all square to $G$, and $TG$ and $PG$ commute because $T,P$ commute. HenceEquivalently, $G^{\prime }$ is the full real $O(1,1)$ group and $G$ its identity component. This is the component quotient of the real Lorentz group in two dimensions; the quotient records the four components, not the continuous boost parameter.