in $A_{\infty}$ such that $$\alpha_{{\infty} g} (x) = w(g)x w(g)^*$$ for all $x \in A$ and all $g \in G$.

The proof is essentially the same as that of Lemma \[Lem.PZ.20200623\].

\[Lem\_inlim\_in\_ap\] Let $G$ be a finite group, let $\big((G,A_j, \alpha^{(j)})_{j \in J} , (\varphi_{j,k})_{j\leq k} \big)$ be a direct system of $G$-algebras, let $A$ be the direct limit of the $A_j$, and let $\alpha \colon G \to {{\operatorname{Aut}}}(A)$ be the direct limit of the $\alpha^{(j)}$. If $\alpha^{(j)}$ is inner for each $j$, then $\alpha$ is strictly approximately inner.

Let ${\varepsilon}> 0$, let $m \in {{\mathbb{Z}}_{> 0}}$, and let $a_1, a_2, \ldots, a_m \in A$. Choose $n \in {{\mathbb{Z}}_{> 0}}$ and $b_1, b_2, \ldots, b_m \in A_n$ such that, for $j = 1, 2, \ldots, m$, $$\label{Eq1.24.07.19}
\| \varphi_{\infty, n} (b_j) - a_j \| < \frac{{\varepsilon}}{2}.$$ Since $\alpha^{(n)}$ is inner, there is a homomorphism $g \mapsto z_g$ from $G$ to ${{\operatorname{U}}}(A_n)$ such that, for all $j\in \{ 1,2,\cdots, n\}$ and all $g \in G$, $$\label{Eq2.24.07.19}
\alpha^{(n)}_g (b_j) = z_g b_j z^*_g.$$ Define $w_g = \varphi_{{\infty}, n} (z_g)$. Therefore, using $\varphi_{\infty, n} \circ \alpha^{(n)}_g = \alpha_g \circ \varphi_{\infty, n}$ and (\[Eq2.24.07.19\]) at the first step and using (\[Eq1.24.07.19\]) at the second 