This is a reference request for something that is likely to be well-known to operator algebraists. I will not, therefore, include the technical definition of free product of finite von Neumann algebras, but instead refer the reader to [Ching - Free products of von Neumann algebras](https://www.ams.org/journals/tran/1973-178-00/S0002-9947-1973-0326405-3) for the definition.


Theorem 3.5 of [Dykema - Interpolated free group factors](https://projecteuclid.org/download/pdf_1/euclid.pjm/1102622631) (letting $A=L(\mathbb{Z})$ and $B=\mathbb{C}$) gives that $M\_{2}(L(\mathbb{\mathbb{Z}}))\*L(\mathbb{Z}\_{2})\cong M\_{2}(L(\mathbb{F}\_{3}))$. Is it known whether or not $L(\mathbb{Z}\*\mathbb{Z}\_{2})\cong L(\mathbb{Z})\*L(\mathbb{Z}\_{2})$ is a free group factor, or interpolated free group factor?


I am, of course, interested in related results like the one quoted above, as well, if the original question is still unknown. Please feel free to provide references as answers.



Is $L(\mathbb{Z}*\mathbb{Z}_{2})$ a free group factor?
This should just be a comment- but for some reason I couldn't add a comment.


It seems to me that using Corollary 5.3 of [this paper](https://arxiv.org/pdf/funct-an/9211013.pdf) by Dykema, we indeed get a positive answer to your question.


Corollary 5.3 states that $L(G \ast H) \cong L(F(2-|G|^{-1}-|H|^{-1}))$, if $G$ and $H$ are nontrivial amenable groups, with $|G|+|H| \geq 5$.( $\infty ^{-1}=0$).


So $L(\mathbb Z \ast \mathbb Z\_2)= L(F(1.5))$ according to the above formula (provided I subtracted correctly).


EDIT: I also found Theorem 1.1 in [this paper](https://projecteuclid.org/download/pdf_1/euclid.pjm/1102785073) to be very interesting. It relates to reduced $C^{\ast}$-algebras of free product groups.


