By polar decomposition, every continuous linear function $f \colon H \to K$ between Hilbert spaces can be written uniquely as $f = \widehat{f} \circ |f|$ for a positive operator $|f| \colon H \to H$ and a partial isometry $\widehat{f} \colon H \to K$ with $\ker(\widehat{f})=\ker(|f|)$. The binary operation $(f,g) \mapsto |g \circ f|$ on the set of positive operators is not associative. Is the binary operation $(f,g) \mapsto \widehat{g \circ f}$ on the set of partial isometries associative? 



Associativity of polar decomposition
Ah, (quite) some fiddling with Mathematica gave a counterexample.
In the notation of anon's answer, take 
$$
 A' = \begin{pmatrix} 1 & 1 & 1 & 1 \\\\ 1 & 1 & 1 & 0 \end{pmatrix}, \qquad
 B = \begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}, \qquad
 C = \begin{pmatrix} 0 & 1/\sqrt{2} \\\\ 1 & 0 \\\\ 0 &
 1/\sqrt{2} \end{pmatrix}.
$$
Then $A=P(A')$, $B$ and $C$ are partial isometries, but $P(P(CB)A) \neq P(CP(BA))$.
Notice that $C$ is a partial isometry that is not an isometry, i.e. has nontrivial kernel.


