Yes, $Q_8 \times \mathbb{Z}_2^\omega$ is balanced, where $Q_8$ is the quaternion group of order $8$ and $\mathbb{Z}_2^\omega$ is the direct sum of infinitely many copies of $\mathbb{Z}_2$.
Yes, $Q_8 \times \mathbb{Z}_2^\omega$ is balanced, where $Q_8$ is the quaternion group of order $8$ and $\mathbb{Z}_2^\omega$ is the direct sum of infinitely many copies of $\mathbb{Z}_2$.