# Nilpotent implies normalizer condition

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 Self-normalizing subgroup (?).

## Proof

### 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, $H \ne N_G(H)$.