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 $G\cong \{0,1\}^{<\omega}$?



Countable abelian group of exponent $2$
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 $\omega$.)


