What is the quotient group of $(\mathbb{Z}/n\mathbb{Z})^* / \{1,-1\}$ ?
If I take the usual decomposition via the Chinese remainder theorem of $(\mathbb{Z}/n\mathbb{Z})^*$ into a direct product of cyclic groups , unfortunately {-1,1} is not contained in any of the factors. I would hope that $(\mathbb{Z}/n\mathbb{Z})^*$ can be decomposed in a slightly different way containing $\{-1,1\} $ as one of the factors, but I am not sure about the details. It might be similar to the composition of the Klein'sche Vierergruppe in 2 different ways, like $V=(e,a)\times(e,b)=(e,ab)\times(e,b)$.
What I hope is to get statements like $(\mathbb{Z}/(4p)\mathbb{Z})^*/\{1,-1\} = C_{p-1}\times C_2/ \{1,-1\} = C_{p-1}$.