Inclusion of G-Hilbert spaces induce homotopy equivalence after stabilization?

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In the paper

https://arxiv.org/pdf/math/0001094.pdf

proposition 6.1, we need to show that the functor $[-]_s$ which sends an equivariant $*$-homomorphism $f$ to its stable homotopy class $[id_{\mathbb{K}(G\mathbb{N})}\otimes f]$ is a stable homotopy functor. It is clear that it is a homotopy functor. What is left is stability, that means for separable $G$-Hilbert spaces $H,H'$ the canonical inclusion $H\hookrightarrow H\oplus H'$ induces a homotopy equivalence $\mathbb{K}(H)\mathbb{K}(G\mathbb{N})A\to \mathbb{K}(H\oplus H')\mathbb{K}(G\mathbb{N})A$.

I know that any separable, infinite dimensional Hilbert space is isomorphic to $l^2(\mathbb{N})$ but I think this does not hold for $G$-Hilbert spaces.

The problem is quite similar to my older questions.

Can somebody help or give a reference?

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I found an answer myself. Actually this is treated in the paper

https://www.jstor.org/stable/pdf/2044632.pdf?refreqid=excelsior%3A48905ad90e43e20fb467a7c9e3310443&ab_segments=&origin=&initiator=

First we note that $A\cong\mathbb{K}(A)$ and $\mathbb{K}(H)\otimes\mathbb{K}(H')\cong \mathbb{K}(H\otimes H')$.

Therefore it suffices to show $G\mathbb{N}\otimes H\otimes A\cong G\mathbb{N}\otimes A$ for all full, countably generated $G$-Hilbert $A$-modules $H$. The desired result then follows for $A=\mathbb{C}$ and that all Hilbert $\mathbb{C}$-modules are full.

Theorem 2.4 of the above paper then yields an isomorphism

$G\mathbb{N}\otimes A=L^2(G)\otimes l^2(\mathbb{N})\otimes A=L^2(G,A)^\infty\cong L^2(G,(A\otimes H))^\infty=G\mathbb{N}\otimes H\otimes A$

which proves the claim.