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Feb 14, 2023 at 13:37 comment added YCor (Typo in the above comment: "The Frobenius" should be "The Frobenius kernel". Of course the Frobenius complement cannot be parameter-free definable since it is not a normal subgroup.)
Feb 8, 2023 at 7:31 vote accept Y. Tamer
Feb 6, 2023 at 14:02 comment added YCor Note: this also shows that the Frobenius kernels and complements are definable (by formulas that don't depend on the Frobenius group). The Frobenius is actually parameter-free definable, namely as the set of those elements whose conjugacy class has a nontrivial centralizer.
Feb 6, 2023 at 13:28 history answered YCor CC BY-SA 4.0