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Locally nilpotent not implies normalizer condition

881 bytes added, 00:24, 17 April 2017
Proof that it does not satisfy the normalizer condition
=== Proof that it does not satisfy the normalizer condition ===
Let <math>H</math> be the subgroup of <math>G</math> generated by the automorphisms: <math>e_i \mapsto e_i + re_j, r \in F, i < j \le 0</math> and: <math>e_i \mapsto e_i + re_j, r \in F, 0 < i < j</math> Basically, we have partitioned the set <math>\mathbb{{fillin}Q}</math> into two totally ordered subsets: the non-positive and the positive numbers, and are only considering the automorphisms in <math>G</math> that keep each part within itself. For an automorphism of the vector space to preserve <math>H</math>, it must either preserve both parts of the partition, or interchange them. However, since <math>G</math> is unitriangular, it cannot send anything in the positive part to anything in the non-positive part. Therefore, it cannot interchange the parts, and therefore it must preserve both parts. We can then show that this forces the normalizing element to be within <math>H</math>.
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