The group von Neumann algebra $L\Gamma$ is a factor if and only if the group $\Gamma$ is ICC (i.e. [infinite conjugacy class property][1]). Moreover if $\Gamma$ is nontrivial then $L\Gamma$ is a $\mathrm{II}_1$ factor. *Question 0.1* (Sakai, Problem 4.4.10 [here][2], 1971): Is every $\mathrm{II}_1$ factor a group von Neumann algebra? *Answer* (Connes [here][3], 1975): No, because there is a $\mathrm{II}_1$ factor which is *not* anti-isomorphic to itself (whereas $L\Gamma$ is so). *Question 0.2*: Is every $\mathrm{II}_1$ factor anti-isomorphic to itself a group von Neumann algebra? *Answer* (Jones, [here][4], 1980): No, because there is a $\mathrm{II}_1$ factor which is anti-isomorphic to itself but without involutory antiautomorphisms (whereas $L\Gamma$ has so). The answers for the next two questions were pointed out by Jiang's comments. *Question 0.3* (Remark 5.7 in previous paper): Is a $\mathrm{II}_1$ factor with an involutory antiautomorphism a group von Neumann algebra? *Answer* (Ioana [here][5], 2010): No, because there are $\mathrm{II}_1$ factors $M$ whose amplifications $M^t$ (with $t \neq 1$) admit an involutory antiautomorphism but are not group von Neumann algebras (see Corollary F, Corollary 10.1 and Remark 10.3). *Question 0.4*: Is a $\mathrm{II}_1$ factor with an involutory antiautomorphism stably isomormphic to a group von Neumann algebra (i.e. of the form $(L\Gamma)^t$)? *Answer* (Boutonnet [here][6], 2013): No, because there are crossed product von Neumann algebra $L^{\infty}(X,\mu) \rtimes_{\sigma} \Gamma$, with $σ$ a free ergodic pmp-action of an ICC group $Γ$ on a probability space $(X,μ)$, which are not stably isomorphic to a group von Neumann algebra. **Question**: Is a $\mathrm{II}_1$ factor with an involutory antiautomorphism stably isomorphic to a crossed product $A \rtimes G$, with $A$ an abelian von Neumann algebra and $G$ a group? Recall that if the action of the group $G$ on the [abelian von Neumann algebra][7] $A$ is free then $A \rtimes G$ is a factor iff the action is ergodic. Now $A \rtimes G$ can be a factor without the action being free, for example when $A = \mathbb{C}$, because then $A \rtimes G = LG$. *Remark*: The existence of a Kac algebra generating a $\mathrm{II}_1$ factor not stably isomorphic to any $A⋊G$ (as above) would be very interesting. The initial question and then Connes' paper were pointed out to me by [Keshab Chandra Bakshi][8]. [1]: https://en.wikipedia.org/wiki/Infinite_conjugacy_class_property [2]: https://mathscinet.ams.org/mathscinet-getitem?mr=0442701 [3]: https://mathscinet.ams.org/mathscinet-getitem?mr=377534 [4]: https://mathscinet.ams.org/mathscinet-getitem?mr=585235 [5]: https://mathscinet.ams.org/mathscinet-getitem?mr=2813341 [6]: https://mathscinet.ams.org/mathscinet-getitem?mr=3077866 [7]: https://en.wikipedia.org/wiki/Abelian_von_Neumann_algebra#Classification [8]: https://mathscinet.ams.org/mathscinet/search/author.html?mrauthid=1197952