Locally nilpotent not implies normalizer condition

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., locally nilpotent group) need not satisfy the second group property (i.e., group satisfying normalizer condition)
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Statement

It is possible to have a group $G$ that is a locally nilpotent group (every finitely generated subgroup of it is a nilpotent group) but is not a group satisfying normalizer condition. In other words, $G$ has a proper subgroup $H$ that is self-normalizing: $H=N_{G}(H)$ .

We can construct examples both for $G$ a torsion-free group and for $G$ a p-group for any prime number $p$ .

Proof

Description of the example

Take a field $F$ . Choose $F$ as follows:

If we want a torsion-free example, take $F=\mathbb {Q}$ , or any other field of characteristic zero.
If we want an example that is a p-group for some prime number $p$ , take $F$ to be the field $\mathbb {F} _{p}$ of $p$ elements, or alternatively, any field of characteristic $p$ .

Let $G$ be McLain's group $M(\mathbb {Q} ,F)$ .

Proof that it is locally nilpotent

$G$ is locally nilpotent because any finitely generated subgroup of $G$ has finite support in the basis $\mathbb {Q}$ , i.e., it moves only finitely many basis elements, so it is in a finite-degree unitriangular matrix group over $F$ , which is nilpotent.

Proof that it does not satisfy the normalizer condition

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References

Textbook references

Book	Page number	Chapter and section	Contextual information	View
A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613^{More info}	369	Section 12.2	proof in exercise 12.2.8	Google Books