# Not measure equivalent ICC groups $G$ and $H$, but $L(G)\cong L(H)$

Some years ago, I heard about the following problem:

Find two ICC, i.e., infinite conjugacy class groups $G$ and $H$, such that $L(G)\cong L(H)$, but $G$ and $H$ are not measure equivalent, where $L(G)$ denotes the group von Neumann algebra associated to the group $G$.

Recall that $G$ and $H$ are measure equivalent iff they admit stably orbit equivalent actions.

What is the status on this problem now?

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The way you stated it, you can take any group $G$ which admits a finite index subgroup $H$ and so that $L(G)$ has some restrictions on the fundamental group. I believe such groups have been constructed. If you ask only for stable equivalence of the von Neumann algebras, I do not think that there are examples. –  Andreas Thom Feb 19 at 22:07
sorry, I do not quite understand the meaning you say "so that L(G) has some restrictions on the fundamental group". The problem asked for constructing $G, H$ such that $L(G)\cong L(H)$ and $G, H$ are not measure equivalent or show such $G, H$ do not exist. Is the expression misleading or am I miss some point? –  Jiang Feb 19 at 22:36
This problem is still open as far as I know. –  Jesse Peterson Feb 21 at 1:08