Is every subgroup of infinite Boolean group finite?

Question

Definition (Boolean group): A group $(G,*)$ is said to be Boolean if every non identity element has order $2$ $i.e$ for any $a\in G,o(a)=2$, where $a\neq e$

Now I want to know that given any infinite Boolean group $(G,*)$, is it true that every proper subgroup of $G$ is finite?

I searched on internet and found that Prüffer $p$ groups are one in which every proper subgroup is finite, but are they only group with this property being infinite, but all proper subgroups are finite)?

Answer 1

Is every subgroup of infinite Boolean group finite?

No.

Consider

$$\prod_{i=1}^\infty\Bbb Z_2.$$

The subgroup $$\prod_{i=2\\ i\text{ even}}^\infty \Bbb Z_2$$ is infinite.

Answer 2

No, and more precisely: every nontrivial Boolean group has a subgroup of index 2. In particular, every infinite Boolean group has infinitely many infinite subgroups.

Proof: let $G$ be such a group and $x$ a nontrivial element of $G$. Let $H$ be a subgroup that is maximal for the property of not containing $x$ (this exists by Zorn's lemma). Since $G$ is abelian, $H$ is normal in $G$. Then the Boolean group $G/H$ has the property that every nontrivial subgroup contains the image $\bar{x}\neq 0$ of $x$. It follows that $G/H=\{0,\bar{x}\}$, so $H$ has index 2. (Finding a subgroup of index 2 in $H$ and so on yields the additional statement.) $\Box$

Notes:

1. the proof works with no change with elementary abelian $p$-groups for prime $p$ (to construct a subgroup of index $p$). Above is the case $p=2$.

2. one can check that a $p$-elementary abelian group of infinite order $\alpha$ has exactly $2^\alpha$ subgroups of index $p$.

3. one can also check that a $p$-elementary abelian group of infinite cardinal $\alpha$ admits $2^\alpha$ subgroups of order $\alpha$ and index $\alpha$.