Why does acting as the inverse on a subspace guarantee commutativity? Let $V$ be some vector space and $P$ a projection onto a subvectorspace $C \subset V$. Let $\mathcal L$ be a set of unitary operators on $V$ satisfying $AP = PAP$ for all $A \in \mathcal L$. In this paper, the authors prove that $\mathcal L$ is a group. The step in their proof which confuses me a bit is when they show that this set contains inverses. Suppose $A \in \mathcal L$. They show
$$(PA^\dagger P)(PAP) = PA^\dagger P^2AP = PA^\dagger (PAP) = PA^\dagger A P = P,$$
which shows that $PA^\dagger P$ is the inverse of $PAP$ on the subspace $C$. They claim that because of this, $P = (PAP)(PA^\dagger P)$, i.e. that $PAP$ and $PA^\dagger P$ commute, and use that fact to complete the proof.
I don't fully understand what they mean when they say that $PA^\dagger P$ is the inverse of $PAP$ on a subspace. I see that $P$ acts as the identity on the subspace, so I can see what they are getting at but I don't see why that guarantees that we can flip around the product. An explanation or proof of this fact would be appreciated!
 A: More generally, let $M$ be a semi-group (i.e., a set with a multiplication that is associative; no neutral or inverse is postulated) and $P\in M$ an element with $P^2=P$. Define $$\mathcal L=\{\,A\in M\mid AP=PAP\,\}.$$
Then

*

*If $A,B\in\mathcal L$ then $ABP=APBP=PAPBP=PABP$, i.e., also $AB\in\mathcal L$

*$P\in\mathcal L$.

Further, for $A,B\in\mathcal L$,
define the equivalence relation $$A\sim B\iff AP=BP$$
and set
$$ \mathcal G=\mathcal L/{\sim}$$
and for $A\in \mathcal L$ let $[A]$ denote the equivalence class of $A$ in $\mathcal G$.
If $A\sim A'$ and $B\sim B'$, then
$ A'B'P=A'PB'P=APBP=ABP$, i.e., $A'B'\sim AB$. Thus multiplication $[A][B]:=[AB]$ in $\mathcal G$ is well-defined and turns $\mathcal G$ into a semi-group.
In this, $[P]$ is right-neutral:
$$ P^2=P\implies APP=AP\implies AP\sim A\implies [A][P]=[A].$$
and also left-neutral:
$$ PA\sim PAP=AP\sim A\implies [P][A]=[A].$$
Moreover, you already know that every element of $\mathcal G$ has a left-inverse, namely for arbitrary $[A]\in\mathcal G$, we have $[A^\dagger][A]=[P]$.
The rest is an abstract and general argument: a semi-group with left-neutrals and left-inverses is a group.
