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Let $\{0,1\}^{<\omega}$ denote the subgroup of $\{0,1\}^{\omega}$ where $f:\omega\to\{0,1\}$ is a member of $\{0,1\}^{<\omega}$ if there is $N\in\omega$ such that $f(n)= 0$ for all $n\in\omega$ with $n\geq N$.

If $(G,+)$ is an abelian group with $|G|=\aleph_0$ and $g+g = 0$ for all $g\in G$, does this imply $G\cong \{0,1\}^{<\omega}$?

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  • $\begingroup$ It is tempting to think so, as one can look at a direct sum of Z2 with index set G, and "remove" those indices c where c=ab and (under some well ordering of G) a and b are before c. However, there may be some exotic behaviour being overlooked by this view point. Gerhard "Should Look Up Boolean Groups" Paseman, 2019.01.06. $\endgroup$
    – Gerhard Paseman
    Commented Jan 7, 2019 at 6:58
  • $\begingroup$ Subduplicate of MathSE post: math.stackexchange.com/questions/1193556/… $\endgroup$
    – YCor
    Commented Jan 7, 2019 at 15:23
  • $\begingroup$ @YCor : Whether this is a duplicate or not depends on whether one believes in AC or not. With AC the question is trivial (the MSE question a little less so, as the exponent is not prime). Without AC the answer given on MSE is false, even for exponent 2. However, in the countable case we don't need AC. A bijection between $G$ and $\mathbb{N}$ induces a well ordering on $G$. Thus we can recursively construct the lexicographically minimal basis of $G$, which induces an isomorphism $G\cong \{0,1\}^{<\omega}$. $\endgroup$
    – Jan-Christoph Schlage-Puchta
    Commented Jan 13, 2019 at 11:43
  • $\begingroup$ @Jan-ChristophSchlage-Puchta there's no reference to AC in the post, which implicitly means that AC is assumed. The answer by GH, which also uses implicitly AC, was accepted by the OP, which confirms this interpretation. (Under the negation of AC, $|G|$ doesn't make sense by the way, so one should say "$G$ is infinite countable" rather than "$|G|=\aleph_0$".) It's a any useful remark that AC is not needed here. $\endgroup$
    – YCor
    Commented Jan 13, 2019 at 11:57
  • $\begingroup$ @YCor: Without AC cardinality becomes incredibly messy, but that doesn't mean that $|G|$ does not make sense. Also whenever one talks about exotic sets, one should not use sloppy language. "Uncountable" is too often used as a synonym "continuum", and AC is assumed even if it is not needed. $\endgroup$
    – Jan-Christoph Schlage-Puchta
    Commented Jan 13, 2019 at 12:08

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Yes. An abelian group $(G,+)$ of exponent $2$ is the same as a module over $\mathbb{Z}/2\mathbb{Z}$, that is, a vector space over the finite field $\mathbb{F}_2$. If $|G|$ is infinite, then the dimension of $G$ over $\mathbb{F}_2$ is $|G|$ by basic cardinal arithmetic, whence $(G,+)$ is isomorphic to a direct sum of $|G|$ copies of $\mathbb{F}_2$. (For the special case $|G|=\omega$ one needs minimal cardinal arithmetic: the dimension cannot be finite, hence it must be $\omega$.)

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    $\begingroup$ It's worth pointing out a group of exponent 2 is always abelian. Edit: I've just noticed the question title mentions abelian groups, but the body doesn't $\endgroup$
    – Wojowu
    Commented Jan 7, 2019 at 8:04
  • $\begingroup$ @Wojowu: Good point. I had this in my subconscious, this is why I added "abelian" to the title :-) $\endgroup$
    – GH from MO
    Commented Jan 7, 2019 at 8:31

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