**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.