Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a group where every non-identity element has order 2.

If |G| is finite then $G$ is isomorphic to the direct product $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots \times \mathbb{Z}_{2}$.

Is the analogous result $G= \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots $.

true for the case |G| is infinite?

share|improve this question
add comment

3 Answers

up vote 25 down vote accepted

Perhaps the best way to look at the problem is to establish the following more precise result:

For a group $G$, the following are equivalent:
(i) Every non-identity element of $G$ has order $2$.
(ii) $G$ is commutative, and there is a unique $\mathbb{Z}/2\mathbb{Z}$-vector space structure on $G$ with the group operation as addition.

I guess you probably already know how to show that if every nonidentity element has order $2$, $G$ is commutative: for all $x,y \in G$, $e = (xy)^2 = xyxy$. Multiplying on the left by $x$ and on the right by $y$ gives $xy = yx$.

Having established the commutativity, it is convenient to write the group law additively. Then there is only one possible $\mathbb{Z}/2\mathbb{Z}$-vector space structure on $G$, since it remains to define a scalar multiplication and of course we need $0 \cdot x = 0, \ 1 \cdot x = x$ for all $x \in G$. But you should check that this actually works: i.e., defines an $\mathbb{Z}/2\mathbb{Z}$-vector space structure, just by checking the axioms: the key point is that for all $x \in G$, $(1+1)x = x + x = 0 = 0x$.

So now your question is equivalent to: is every $\mathbb{Z}/2\mathbb{Z}$ vector space isomorphic to a product of copies of $\mathbb{Z}/2\mathbb{Z}$? Well, the only invariant of a vector space is its dimension. It is clear that every finite-dimensional vector space is of this form. Every infinite dimensional space is isomorphic to a direct sum $\bigoplus_{i \in I} \mathbb{Z}/2\mathbb{Z}$, the distinction being that in a direct sum, every element has only finitely many nonzero entries. (In other words, the allowable linear combinations of basis elements are finite linear combinations.) Moreover, for any infinite index set $I$, the direct sum $\bigoplus_{i \in I} \mathbb{Z}/2\mathbb{Z}$ has dimension $I$ and also cardinality $I$.

Finally, it is not possible for a direct product of two element sets to have countably infinite cardinality: if $I$ is infinite, it is at least countable, and then the infinite direct product has the same cardinality of the real numbers (think of binary expansions). So the answer to your question is "yes" for direct sums, but "no" for direct products.

share|improve this answer
    
I think you have a small typo in the third paragraph, where you have $1 \cdot x = 1$, I believe you should have $1 \cdot x = x$. By the way, really nice answer =) –  Adrián Barquero Jan 21 '11 at 3:52
    
@Adrián: I fixed it; thanks. –  Pete L. Clark Jan 21 '11 at 5:07
add comment

If every non-identity element of $G$ has order 2, then the group is abelian. Notationally, it helps to write the group operation additively, with identity $0$. In that case, you can view the group as a vector space over the field with two elements, $F_2=\mathbb{Z}/2\mathbb{Z}$. Every vector space has a basis, so $$ G\cong\bigoplus_{i\in I}F_2 $$ where $I$ has the cardinality of a basis of $G$. The point here is that representing a vector space by a basis corresponds to a direct sum rather than the direct product, because you can only take finite linear combinations of basis elements. (and, as pointed out in Andres' answer, it is not possible to represent all such groups as direct products).

share|improve this answer
add comment

Not really. The direct sum $\bigoplus_{n\in{\mathbb N}}{\mathbb Z}_2$ is a counterexample. More generally, take $G=\prod_{n=1}^\infty{\mathbb Z}_2$. This set has size $|{\mathbb R}|$. It is easy to see that, given any element of this group, there is a countable subgroup of $G$ that contains this element. Much more pathological counterexamples are also possible, of course.

share|improve this answer
    
The countable subgroup of $G$ which contains the element can be taken to be the cyclic group it generates :) –  Mariano Suárez-Alvarez Jan 11 '11 at 1:51
    
:-) I meant countably infinite. Anyway, we can in fact create a strictly increasing sequence of length $\omega_1$ of countable elementary subgroups of $G$. (Elementary in the "elementary substructure" sense from mathematical logic.) –  Andres Caicedo Jan 11 '11 at 1:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.