Show that $p \leq q \iff pq = qp = p.$

Question

Let $p,q$ be two projections in a $C^{\ast}$-algebra $A.$ Then show that $$p \leq q \iff pq = qp = p.$$

By projection we usually mean those elements in a $C^{\ast}$-algebra which are self-adjoint and idempotent. With this definition in mind how do I approach the problem? Could anyone give me some hints?

Thanks for your time.

Answer 1

(note that assuming $A$ unital changes nothing, and it simplifies arguments)

If $pq=qp=p$, then $$ p=qp=qpq\leq q1q=q^2=q. $$ Conversely, suppose that $p\leq q$. Then $$ 0\leq (1-q)p(1-q)\leq(1-q)q(1-q)=0. $$ So $$(1-q)p(1-q)=0.$$ This we can write as $$ 0=(1-q)p(1-q)=(1-q)p^*p(1-q)=[p(1-q)]^*p(1-q). $$ Thus (thank God for the C$^*$-identity!) $$ 0=p(1-q). $$ That is, $p=pq$. Taking adjoints, $p=qp$.