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Given a finite Von Neumann algebra $(N,\tau)$, and Von Neumann subalgebras $A\subset B$ with the same identity, I came across the fact saying that $B$ has an amenable direct summand implies $A$ has an amenable direct summand.

At first glance, this is easy to proof, since say $Bp$ is amenable for some nonzero central projection $p$ in $B$, then of course, $Ap$ is amenable, but $p$ is not necessarily in $A$, of course, $q=E_A(p)$ lies in center of $A$, but it is not a projection anymore, and I do not know whether $Aq$ is amenable, intuitively, since $q$ is a positive operator, we can take a nonzero spectral projection $r$ of $q$, with $r$ lying in the center of $A$, and try to show $Ar$ is amenable, but I am not able to fill the details, can anyone help me with this?

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What is $B$, and how does it relate to $A$ and $N$? –  Owen Sizemore Jul 26 at 21:45
Oops, sorry, I should have mentioned $A\subset B\subset N$. –  ougao Jul 26 at 22:35

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