# Finite subgroup containing all finite subgroups of infinite group

The proof of the following

Theorem In an infinite 2-group every finite subgroup is properly contained in its normalizer.

begins with

Let $$F$$ be a finite subgroup of an infinite 2-group $$G$$ and assume that $$F = N_G(F)$$. Since $$G$$ is infinite, not all its finite subgroups are contained in $$F$$.

I know that an infinite group has got infinitely many subgroups, but it's not true that it has got infinitely many finite subgroups, so I don't see why the sentence "not all its finite subgroups are contained in $$F$$" is true. For example, $$(\mathbb{Z}, +)$$ is an infinite group but it has got only infinite subgroups.

I think I should use some property of 2-groups (or maybe the normalizer?).

• References: The theorem is from "Finiteness Conditions and Generalized Soluble Groups - Part 1" by Derek J. S. Robinson
• Take an element $g \in G - F$. The conjugate $g^{-1}Fg$ is a finite subgroup that's not contained in $F$ since $g$ lies outside of the normalizer of $F$. Commented Sep 17, 2022 at 18:38

In a $$p$$-group, every element has $$p$$-power order, so every element has finite order. Since every element is in the cyclic subgroup it generates, every element is in a finite subgroup. If a subgroup contained every finite subgroup, it would then contain every element, i.e. be the whole group.