[proofplan] We prove that the center of $L(G)$ consists only of scalar multiples of the identity operator. For a central element $x\in L(G)$, we examine the Fourier coefficient function $g\mapsto (x\delta_e,\delta_g)_{\ell^2(G)}$. Centrality forces this coefficient function to be constant on conjugacy classes, while square-summability forces it to vanish on every infinite [conjugacy class](/page/Conjugacy%20Class). The ICC hypothesis therefore leaves only the identity coefficient, and the standard separating property of the canonical vector $\delta_e$ for $L(G)$ then implies that $x$ is scalar. [/proofplan] [step:Fix the regular representations and the Fourier coefficients of a central element] Let $e\in G$ denote the identity element. Let \begin{align*} \lambda:G\to \mathcal{L}(\ell^2(G)) \end{align*} be the left regular representation, defined on the canonical [orthonormal basis](/page/Orthonormal%20Basis) $(\delta_g)_{g\in G}$ by $\lambda_h\delta_g=\delta_{hg}$ for $g,h\in G$. Let \begin{align*} \rho:G\to \mathcal{L}(\ell^2(G)) \end{align*} be the right regular representation, defined by $\rho_h\delta_g=\delta_{g h^{-1}}$ for $g,h\in G$. By the definition fixed in the theorem statement, $L(G)=\rho(G)'$. Define the center of $L(G)$ by \begin{align*} Z(L(G))=\{z\in L(G):zy=yz\text{ for every }y\in L(G)\}. \end{align*} Let $x\in Z(L(G))$. For each $g\in G$, define the Fourier coefficient \begin{align*} a_g=(x\delta_e,\delta_g)_{\ell^2(G)}. \end{align*} Since $x\delta_e\in \ell^2(G)$, the coefficient family $(a_g)_{g\in G}$ is square-summable: \begin{align*} \sum_{g\in G}|a_g|^2=\|x\delta_e\|_{\ell^2(G)}^2<\infty. \end{align*} [/step] [step:Use centrality to show that the Fourier coefficients are constant on conjugacy classes] Fix $g,h\in G$. Since $x\in Z(L(G))$ and $\lambda_h\in L(G)$, we have $x\lambda_h=\lambda_h x$. Also $x\in L(G)=\rho(G)'$, so $x\rho_h=\rho_h x$. Using $\delta_{hgh^{-1}}=\lambda_h\delta_{g h^{-1}}$ and the unitarity of $\lambda_h$, we compute \begin{align*} a_{hgh^{-1}}=(x\delta_e,\delta_{hgh^{-1}})_{\ell^2(G)}. \end{align*} Thus \begin{align*} a_{hgh^{-1}}=(\lambda_h^*x\delta_e,\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} Because $x$ commutes with $\lambda_h$, this becomes \begin{align*} a_{hgh^{-1}}=(x\lambda_h^*\delta_e,\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} Since $\lambda_h^*\delta_e=\delta_{h^{-1}}=\rho_h\delta_e$, we get \begin{align*} a_{hgh^{-1}}=(x\rho_h\delta_e,\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} Since $x$ commutes with $\rho_h$, this is \begin{align*} a_{hgh^{-1}}=(\rho_h x\delta_e,\rho_h\delta_g)_{\ell^2(G)}. \end{align*} The operator $\rho_h$ is unitary, so \begin{align*} a_{hgh^{-1}}=(x\delta_e,\delta_g)_{\ell^2(G)}=a_g. \end{align*} Therefore $a_g=a_{hgh^{-1}}$ for all $g,h\in G$. [guided] We want to extract information from the assumption that $x$ is central. The natural quantities attached to $x$ are its Fourier coefficients \begin{align*} a_g=(x\delta_e,\delta_g)_{\ell^2(G)}. \end{align*} The goal is to prove that these coefficients cannot distinguish conjugate group elements. Fix $g,h\in G$. The vector $\delta_{hgh^{-1}}$ can be written as $\lambda_h\delta_{g h^{-1}}$, because $\lambda_h$ sends $\delta_{g h^{-1}}$ to $\delta_{hgh^{-1}}$. Hence \begin{align*} a_{hgh^{-1}}=(x\delta_e,\lambda_h\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} Since $\lambda_h$ is unitary, moving it to the first variable gives \begin{align*} a_{hgh^{-1}}=(\lambda_h^*x\delta_e,\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} Now centrality is used: because $x\in Z(L(G))$ and $\lambda_h\in L(G)$, we have $\lambda_h^*x=x\lambda_h^*$. Therefore \begin{align*} a_{hgh^{-1}}=(x\lambda_h^*\delta_e,\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} But $\lambda_h^*\delta_e=\delta_{h^{-1}}$, and this same vector is $\rho_h\delta_e$ by the definition $\rho_h\delta_k=\delta_{k h^{-1}}$. Thus \begin{align*} a_{hgh^{-1}}=(x\rho_h\delta_e,\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} The standard commutation relation for the group von Neumann algebra says $L(G)=\rho(G)'$, so every element of $L(G)$ commutes with every $\rho_h$. Since $x\in L(G)$, we may rewrite this as \begin{align*} a_{hgh^{-1}}=(\rho_h x\delta_e,\delta_{g h^{-1}})_{\ell^2(G)}. \end{align*} Finally, $\delta_{g h^{-1}}=\rho_h\delta_g$, and $\rho_h$ is unitary, so \begin{align*} a_{hgh^{-1}}=(\rho_h x\delta_e,\rho_h\delta_g)_{\ell^2(G)}=(x\delta_e,\delta_g)_{\ell^2(G)}=a_g. \end{align*} This proves that the Fourier coefficient function $g\mapsto a_g$ is constant on each conjugacy class. [/guided] [/step] [step:Use the ICC hypothesis and square summability to kill all nonidentity coefficients] Let $g\in G$ with $g\ne e$. Define its conjugacy class by \begin{align*} \mathcal{C}(g)=\{hgh^{-1}:h\in G\}. \end{align*} By the ICC hypothesis, $\mathcal{C}(g)$ is infinite. From the previous step, $a_k=a_g$ for every $k\in \mathcal{C}(g)$. Since $(a_k)_{k\in G}\in \ell^2(G)$, we have \begin{align*} \sum_{k\in \mathcal{C}(g)}|a_k|^2\le \sum_{k\in G}|a_k|^2<\infty. \end{align*} If $a_g\ne 0$, then the left-hand side is the sum of infinitely many equal positive terms $|a_g|^2$, hence is infinite. This contradiction gives $a_g=0$. Therefore \begin{align*} a_g=0 \end{align*} for every $g\in G\setminus\{e\}$. [/step] [step:Apply separation of the canonical vector to conclude that the central element is scalar] The preceding step shows that \begin{align*} x\delta_e=a_e\delta_e. \end{align*} Let $I_{\ell^2(G)}:\ell^2(G)\to\ell^2(G)$ denote the identity operator, so $I_{\ell^2(G)}\xi=\xi$ for every $\xi\in\ell^2(G)$. Let \begin{align*} y=x-a_e I_{\ell^2(G)}\in L(G). \end{align*} Then $y\delta_e=0$. Since $L(G)=\rho(G)'$, the operator $y$ commutes with every $\rho_h$. For each $g\in G$, using $\delta_g=\rho_{g^{-1}}\delta_e$, we obtain \begin{align*} y\delta_g=y\rho_{g^{-1}}\delta_e=\rho_{g^{-1}}y\delta_e=0. \end{align*} The linear span of $(\delta_g)_{g\in G}$ is dense in $\ell^2(G)$, and $y$ is bounded, so $y=0$. Hence \begin{align*} x=a_e I_{\ell^2(G)}. \end{align*} Thus every element of $Z(L(G))$ is a scalar multiple of the identity. Since $L(G)$ is a von Neumann algebra and its center is $\mathbb C I_{\ell^2(G)}$, $L(G)$ is a factor. [/step]