Prove that Prove that $Q_{8}'=\{1,-1\}$.
My proof:
$Q_{8}'\neq \{1\}$, because $Q_{8}$ is not abelian. $|Q_{8}/\{-1,1\}|=4$. So $Q_{8}/\{-1,1\} \cong \mathbb{Z}_4$ or $Q_{8}/\{-1,1\} \cong \mathbb{Z}_2\times\mathbb{Z}_2$. Then $Q_{8}/\{-1,1\}$ is abelian.
Then I use a theorem:
Let $N \unlhd G $. Then $G/N$ is abelian $\Leftrightarrow G'\subseteq N$.
Then $Q_{8}'\subseteq\{-1,1\}$. But we know that $Q_8'$ can't be $\{1\}$. Then $Q_{8}'=\{-1,1\}.$
Is it correct?