= Solution
The <subgroup> generated by $T,P$ is $H=\{I,T,P,-I\}$, a <Klein four-group>. It is not normal in $G'$. Indeed,
$$
B(x)TB(-x)=B(2x)T=\begin{pmatrix}\cosh2x&\sinh2x\\-\sinh2x&-\cosh2x\end{pmatrix}.
$$
For $x\ne0$ the off-diagonal entries are nonzero, whereas every member of $H$ is diagonal. Thus this conjugate of $T$ is outside $H$, giving
$$
\boxed{\langle T,P\rangle\not\trianglelefteq G'.}
$$
The <normal subgroup> in part (c) is the continuous boost <subgroup>; choosing one representative from each of its <cosets> does not make those representatives a <normal subgroup>.
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