I want to prove that $(\mathbb{Z}/3\mathbb{Z})^\times/((\mathbb{Z}/3\mathbb{Z})^\times)^2$ is isomorphic to $\{\pm1\}$.
What I'm having trouble seeing is what the elements of $G=(\mathbb{Z}/3\mathbb{Z})^\times/((\mathbb{Z}/3\mathbb{Z})^\times)^2$ look like. When one writes $a \cdot ((\mathbb{Z}/3\mathbb{Z})^\times)^2$, $a\in (\mathbb{Z}/3\mathbb{Z})^\times$, do they mean $(a,1) \cdot ((\mathbb{Z}/3\mathbb{Z})^\times)^2$?
Also, am I correct in thinking that $(\mathbb{Z}/3\mathbb{Z})^\times=\{1,2\}$ is exactly the group $\{1,-1\}$?