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 a group with $|G|=\aleph_0$ and $g+g = 0$ for all $g\in G$, does this imply $G\cong \{0,1\}^{<\omega}$?

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

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 a group with $|G|=\aleph_0$ and $x+x = 0$ for all $g\in G$, does this imply $G\cong \{0,1\}^{<\omega}$?

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 a group with $|G|=\aleph_0$ and $g+g = 0$ for all $g\in G$, does this imply $G\cong \{0,1\}^{<\omega}$?

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 a group with $|G|=\aleph_0$ and $x+x = 0$ for all $g\in G$, does this imply $G\cong \{0,1\}^{<\omega}$?

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 a group with $|G|=\aleph_0$ and $x+x = 0$ for all $g\in G$, does this imply $G\cong \{0,1\}^{<\omega}$?