**Question:** Let $G$ be a finite group and let $P$ be a $\rm II\_1$ factor. Assume that $G$ acts on $P$ in a trace-preserving manner, such that the crossed product algebra $P \rtimes G$ is a factor. Is $G \curvearrowright^{\sigma} P$ outer?


**Motivation:** It's mentioned in Example 2.3.3(b) in Jones-Sunder's book that the above result is true. However, the proof is not given. I tried to prove it myself, but got stuck. Below is my attempt.


**Attempt:** Suppose $G \curvearrowright^{\sigma} P$ is not outer. Then there exists $g \in G$, $g \neq e$ such that $\sigma\_g$ is inner. Assume that $\sigma\_g =id$. Let $v =\sum\_{h \in G} u\_h u\_g u\_{h^{-1}}$. Then, it's easy to verify that $v $ lies in the center of $P \rtimes G$, and hence is a scalar. Now, let $H=C\_G(g)$. Then, $v= \sum\_{h \in H} u\_g + \sum\_{h \in G \setminus H} u\_{hgh^{-1}}=c \in \mathbb C$. Therefore, $|H| + \sum\_{h \in G \setminus H} u\_{hgh^{-1}g^{-1}}=cu\_{g^{-1}} $. Taking traces on both sides, we get that $|H|=0$, which is a contradiction.


I tried a similar trick when $\sigma\_g=ad(u)$ for some unitary $u \in P$. I took $v= \sum\_{h \in G} u\_h u^{\ast}u\_{gh^{-1}}$. I can show that $v $ lies in the center of $P \rtimes G$, and hence is a scalar. However, I got stuck after that.



Action of a finite group on a finite factor
I think that the claim in the question is false. One can construct a counterexample as follows. First assume in general that $G$ is a finite abelian group of order $n$ and that $\Omega : G \times G \to S^1$ is a bicharacter (i.e. a map that is multiplicative in both variables). Define the projective representation $U : G \to U(\ell^2(G))$ by
$$U(g)e\_k = \overline{\Omega(g,k)} e\_{gk}$$
where $(e\_k)\_{k \in G}$ is the standard orthonormal basis of $G$. One checks that
$$U(g) U(h) = \overline{\Omega(g,h)} U(gh)$$
for all $g,h \in G$. Identify $\ell^2(G) \cong \mathbb{C}^n$ and let $P\_0$ be any $II\_1$ factor. Define $P = M\_n(\mathbb{C}) \otimes P\_0$ and the action $G \curvearrowright^\sigma P$ by $\sigma\_g = \operatorname{Ad}(U(g) \otimes 1)$. Then every automorphism $\sigma\_g$ is inner by construction.


Consider the twisted group von Neumann algebra $L\_\Omega(G)$ generated by canonical unitary operators $(v\_g)\_{g \in G}$ satisfying
$$v\_g v\_h = \Omega(g,h) v\_{gh} \; .$$
There then is a $\*$-isomorphism
$$\pi : P \rtimes\_\sigma G \to P \otimes L\_\Omega(G) : \pi(a u\_g) = a(U(g) \otimes 1) \otimes v\_g$$
for all $a \in P$ and $g \in G$.


To produce a counterexample, it now suffices to give an example where $L\_\Omega(G) \cong M\_n(\mathbb{C})$. This happens for instance when $\Gamma$ is a finite abelian group and $G = \Gamma \times \widehat{\Gamma}$ with
$$\Omega : G \times G \to S^1 : \Omega((g,\omega),(h,\eta)) = \omega(h) \; .$$
Defining the unitary operators $W(g,\omega)$ on $\ell^2(\Gamma)$ by
$$W(g,\omega)e\_h = \omega(h) e\_{gh} \; ,$$
it follows that
$$\theta : L\_\Omega(G) \to B(\ell^2(\Gamma)) \cong M\_n(\mathbb{C}) : \theta(v\_{(g,\omega)}) = W(g,\omega)$$
is a $\*$-isomorphism.


So for instance the group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ admits an inner action on a $II\_1$ factor such that the crossed product is a factor.


