# Is every subgroup of infinite Boolean group finite?

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)?

• How about $\mathbb{Z}_{2}^{\mathbb{N}}$ Dec 25, 2021 at 7:35
• What about $\bigoplus_{k\in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$, it is clearly boolean and it has an infinite subgroup isomorphic to itself. Dec 25, 2021 at 7:35
• Ahh, yes,it is......what about group of power set of infinite set,with binary operation being symmetric difference on sets? Dec 25, 2021 at 7:39

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.

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 $$\alpha$$.