Nilpotent implies normalizer condition

Statement
Any fact about::nilpotent group satisfies the normalizer condition: it has no proper fact about::self-normalizing subgroup.

Facts used

 * 1) uses::Nilpotent implies every subgroup is subnormal
 * 2) uses::Every subgroup is subnormal implies normalizer condition

Proof using given facts
The proof follows directly from facts (1) and (2).

Direct proof
Given: A nilpotent group $$G$$, a proper subgroup $$H$$ of $$G$$. $$N_G(H)$$ denotes the normalizer of $$H$$ in $$G$$.

To prove: $$H$$ is properly contained in $$N_G(H)$$

Proof: We prove this by induction on the nilpotency class of $$G$$. For now, let us assume that the result has been proved for all groups of nilpotency class strictly less than that of $$G$$. The base case of the induction is Abelian groups, where the statement is clearly true (because every subgroup is normal).

Let $$Z(G)$$ denote the center of $$G$$. Clearly $$Z(G) \le N_G(H)$$. Consider two cases:


 * $$Z(G)$$ is not contained in $$H$$: In this case, $$H$$ cannot be equal to $$N_G(H)$$
 * $$Z(G)$$ is contained in $$H$$: In this case, consider the subgroup $$H/Z(G)$$ in the group $$G/Z(G)$$. The nilpotency class of $$G/Z(G)$$ is one less than that of $$G$$, and $$H/Z(G)$$ is a proper subgroup since $$H$$ is proper in $$G$$. Hence, we have:

$$H/Z(G) \ne N_{G/Z(G)}(H/Z(G)) $$

But the right side is the same as:

$$(N_G(H))/(Z(G))$$

showing that $$H/Z(G) \ne N_G(H)/Z(G)$$. Thus, $$H \ne N_G(H)$$.