The left regular factor of an ICC discrete group is a non-type-I factor

The vector functional $\tau$ is positive and unital, since $\tau (T^{\ast }T)=\mathrm{\parallel }T{\delta }_e{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }{\delta }_e\mathrm{\parallel }=1$, and is WOT-continuous, hence normal. On generators, $\tau (\lambda (g)\lambda (h))={\mathbf{1}}_{gh=e}={\mathbf{1}}_{hg=e}=\tau (\lambda (h)\lambda (g))$. Bilinearity proves the trace identity on their linear span. For fixed $A$ in that span, approximate $B\in M$ in WOT and use separate multiplication continuity to obtain $\tau (AB)=\tau (BA)$; then fix this $B$ and approximate arbitrary $A\in M$, proving traciality on $M$. Every $T\in M$ commutes with $\rho (\mathrm{\Gamma })$ by [F1,F2]. If $\tau (T^{\ast }T)=0$, then $T{\delta }_e=0$ and $T{\delta }_g=T\rho (g^{-1}){\delta }_e=\rho (g^{-1})T{\delta }_e=0$ for all $g$, so $T=0$ on a dense basis. Thus $\tau$ is faithful. The operators $\lambda (g)$ are linearly independent: applying a finite linear relation to ${\delta }_e$ gives the corresponding relation among the distinct basis vectors ${\delta }_g$. Hence $M$ is infinite dimensional.

Let $T\in Z(M)$ and write $T{\delta }_e={\sum }_g{c}_g{\delta }_g$. For $s\in \mathrm{\Gamma }$ the unitary $C_s=\lambda (s)\rho (s)$ fixes ${\delta }_e$ and sends ${\delta }_g$ to ${\delta }_{sg{s}^{-1}}$. Centrality and [F1,F2] imply that $T$ commutes with both factors of $C_s$, so $C_{s}T{\delta }_e=T{\delta }_e$. Thus $c_g$ is constant on each conjugacy class. An ${\mathrm{\ell }}^2$ sequence cannot have a nonzero constant value on an infinite set, so ICC gives $T{\delta }_e=c_e{\delta }_e$. Commutation with $\rho$ then gives $T{\delta }_g=c_e{\delta }_g$ for every $g$. Therefore $T=c_{e}I$, proving the factor assertion.

If $M$ were type I, [F4] would give the spatial algebra $\mathcal{B}(E)\otimes I_L$. Finite-dimensional $E$ would make $M$ finite dimensional. If $E$ is infinite dimensional and separable, choose a countable orthonormal basis and the isometries $V_1,V_2$ onto its even and odd basis subspaces. Their range projections $p,q$ are orthogonal. Transferring $V_i\otimes I_L$ to $M$, traciality gives $\tau (p)=\tau (V_1^{\ast }{V}_1)=1$ and $\tau (q)=1$, while positivity and $p+q\le I$ give $2=\tau (p+q)\le 1$, a contradiction. Thus $M$ is not type I and [F4] excludes an irreducible multiple. In particular $\lambda$ is not irreducible. Its one-point integral is a central factor decomposition, since diagonal operators on that point are $\mathbb{C}I=Z(M)$; any central diagonal model must have a one-atom measure algebra on its effective support, because its diagonal algebra is scalar. The fibre therefore remains this non-irreducible factor, rather than an irreducible.

For completeness, $F_2$ is countable by its finite reduced words and infinite by the powers of $a$. If $w\mathrm{\notin }A$, all conjugates $a^{n}w{a}^{-n}$ are distinct: equality at two different integers would make $w$ commute with a nonzero power $a^k$. Then $A\cap w^{-1}Aw$ contains $⟨a^k⟩$, of finite index in both infinite cyclic groups, contradicting ${\mathrm{Comm}}_{F_2}(A)=A$ from [F5]. If $w\in A\setminus \{e\}$, then $w\mathrm{\notin }B$ by [F5], and the same argument with $b$ gives infinitely many conjugates. Thus $F_2$ is ICC. The finite-support permutation group is a countable union of finite permutation groups and is infinite. For a nonidentity permutation with finite moved support $S$, move $S$ to infinitely many pairwise disjoint blocks of the same size by finite permutations. Its conjugates then have distinct moved supports and are distinct, proving ICC. Both examples therefore satisfy all conclusions above.