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
  • 2
    $\begingroup$ 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 $F$. $\endgroup$
    – Chris
    Sep 17, 2022 at 18:38

1 Answer 1


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.


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