Condition for reducible representation So I was reading Georgi's Lie Algebras in Particle Physics. On page 5, he said:

A representation is reducible if it has an invariant subspace, which means that the action of any $D(g)$ on any vector in the subspace is still in the subspace. In terms of the projection operator $P$ onto the subspace, this condition can be written as $$P D ( g ) P = D ( g ) P \forall g \in G$$

I do understand the intuition of reducible representation (I think it means the representation is, well, still a representation inside the subspace alone). But what I don't understand is the condition written above. 
I think it should've said something like: after acting $D(g)$ on the projected vector $P|a>$, $P'|a>=D ( g ) P |a>$ is still in the subspace $S$, right? So I can formulate it as $D ( g ) P |a> \in S, \forall g \in G$, with $|a>$ be an arbitrary vector that these operators on, and $S$ be the vector space of those that are projected vectors $P|a>$, in other words $S \ni P|a>$?  Why is the condition written like that? Thank you!
 A: Let $V$ be the whole space and let $W$ be the subspace. If $P$ is a projection from $V$ onto $W$, then$$\{P(v)\,|\,v\in V\}=W.$$So, the assertion$$(\forall v\in V)(\forall g\in G):P\bigl(D(g)P(v)\bigr)=D(g)P(v)$$is equivalent to$$(\forall w\in W)(\forall g\in G):P\bigl(D(g)w\bigr)=D(g)w.\tag1$$But $P$ is a projection onto $W$ and therefore the assertion $(1)$ simply means that$$(\forall g\in G):w\in W\implies D(g)w\in W.$$In other words, it means that $W$ is an invariant subspace.
A: Your understanding of reducible representations is correct. The issue is with the role of the projection operator.
It is true that they are saying "For every $|a\rangle$ and every $g$ we have that $D(g)P|a\rangle \in S$ where $S$ is the subspace onto which $P$ is the projection.
What they then use is that the statement $|x\rangle \in S$ is equivalent to the statement $|x\rangle = P|x\rangle$. Think about it: for every $|x\rangle \in S$ the operator $P$ just leaves it as it was, and for every $|x\rangle \not\in S$ the operator $P$ maps it to a different vector. So writing $P|x\rangle = |x\rangle$ is the same as writing $|x\rangle \in S$ and that is what they do for the special case of $|x\rangle = D(g)P|a\rangle$. 
