Nilpotent implies normalizer condition
From Groupprops
This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property must also satisfy the second group property
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Statement
Any Nilpotent group (?) satisfies the normalizer condition: it has no proper Selfnormalizing subgroup (?).
Facts used
 Nilpotent implies every subgroup is subnormal
 Every subgroup is subnormal implies normalizer condition
Proof
Proof using given facts
The proof follows directly from facts (1) and (2).
Direct proof
Given: A nilpotent group , a proper subgroup of . denotes the normalizer of in .
To prove: is properly contained in
Proof: We prove this by induction on the nilpotency class of . For now, let us assume that the result has been proved for all groups of nilpotency class strictly less than that of . The base case of the induction is Abelian groups, where the statement is clearly true (because every subgroup is normal).
Let denote the center of . Clearly . Consider two cases:
 is not contained in : In this case, cannot be equal to
 is contained in : In this case, consider the subgroup in the group . The nilpotency class of is one less than that of , and is a proper subgroup since is proper in . Hence, we have:
But the right side is the same as:
showing that . Thus, .