It's well known that two self-adjoint commuting operators have a common set of eigenvectors (see, for example, here). But... isn't it always the case?
Isn't the composition of two self-adjoint operators self-adjoint? If so, then they should always commute, for $AB=(AB)^*=B^*A^*=BA$.
The composition of two self-adjoint operators might not be self-adjoint. For instance, over $V=\mathbb{R}^2$ with the usual inner product, consider $A=\begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$ and $B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, which satisfy $$ AB=\begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix}, $$ which is not self-adjoint.