Decompose ${\mathbb{Z}_n}^{*} / ({\mathbb{Z}_n}^{*})^2$ into a direct sum of cyclic groups,
Where $({\mathbb{Z}_n}^{*})^2 = \left\{a^2 \ : \ a \in {\mathbb{Z}_n}^{*} \right\}$,
A) when $n$ is an odd prime $p$,
B) when $n$ is a product of two distinct prime $p$, and $q$.
I think that they are just $\mathbb{Z}_2$.
Since, if $n=p$ an odd prime, ${\mathbb{Z}_p}^{*} = \left\{ 1, \ 2, \ ... , \ p-1\right\}$, then $({\mathbb{Z}_p}^{*})^2 = \left\{ 1^2, \ 2^2, \ ... . \ (\frac{p-1}{2})^2 \right\}$.
Then ${\mathbb{Z}_p}^{*} / ({\mathbb{Z}_p}^{*})^2 \cong \mathbb{Z} _2$
Similarly, if $n = pq$ for distinct prime $p$ and $q$, ${\mathbb{Z}_{pq}}^{*} = \mathbb{Z}_{(p-1)(q-1)}$.
And if $m \in \mathbb{Z}_{pq}^{*}$, then should $pq-m$ be too.
Clearly, their squares modulo $pq$ is exactly same. So $(\mathbb{Z}_{pq}^{*})^2$ should be $ \mathbb{Z}_{\frac{(p-1)(q-1)}{2}}$.
So is this case, $\mathbb{Z}_{pq}^{*} / (\mathbb{Z}_{pq}^{*})^2 \cong \mathbb{Z}_2$.
It seems to me, then, no matter what $n$ is, $\mathbb{Z}_{n}^{*} / (\mathbb{Z}_{n}^{*})^2 $ is congruent to $\mathbb{Z}_2$.
Where have I done wrong thing? What is wrong?