In induced von Neumann algebra, do we have $P(PMP) =P\cdot P(M)P$ and $U(PMP) =P\cdot U(M)P$? Let $M$ be a von Neumann algebra and $P\in M'$ be a projection. Then, $PMP$ is an induced von Neumann algebra. I know that $ P Z(M) P = Z(PMP)$ and $ (PMP)' = PM'P $ (here, $Z(M)$ is the center and $M'$ is the commutant). I would like to ask, do we have 
$$P(PMP) =P\cdot P(M)P$$
and 
$$U(PMP) =P\cdot U(M)P,$$
where $P(PMP)$ denotes the set of all projections in $PMP$ and $U(PMP)$ denotes all unitary elements of $PMP$.
It looks that Kadison-Ringrose I, Proposition 5.5.5 implies what I want above.

Since $P\in M'$, there is a central carrier $C_p$ of $P$ in $Z(M)$. By  Kadison-Ringrose I, Proposition 5.5.5, for every projection $PQP$ in $PMP$, we have 
$$  QC_p =i(PQP) = i(PQP)i(PQP) = Q C_p QC_p ,$$
where $i$ is the *-isomorphism in that proposition.
Since $C_p$ lies in the central of $M$, it follows that $QC_p +1 -C_p \in M$ and it is a projection.
Similar to the unitary case.
Together with Martin's answer, the question is solved.
