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Nov 21, 2013 at 5:15 comment added Caleb Eckhardt Hi Ulrich, This probably fits in the "stupid answer" category but you could take $\mathbb{R}^{2n}$ many tensor copies of $M_2(\mathbb{C})$ as $A$ then the permutation group of $\mathbb{R}^{2n}$ is contained in $Aut(A)$ by just permuting the tensor factors.
S Nov 20, 2013 at 0:11 history suggested Tobias Fritz
added missing tag
Nov 19, 2013 at 23:56 review Suggested edits
S Nov 20, 2013 at 0:11
Nov 19, 2013 at 23:35 history asked Ulrich Pennig CC BY-SA 3.0