In Dummit and foote book it is written that for any $n \ge 3$, $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ is not cyclic
I understand the proof of $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ is not cyclic for $n \ge3$
Bu i think $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ is also not cyclic for $n \ge2$
For $n=2$
Take the subgroups generated by $2^n-1$ and $2^{n-1} +1$
Now put $ n=2$ then $ 2^2-1=3\equiv1 \mod(2^2) \implies3^2=9=1 \mod4$
similarly $2^{2-1} +1 =3\equiv1 \mod(2^2) \implies3^2=9=1 \mod4$
Both element are distinct element of $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ and have same order $2$
Therefore for any $n \ge 2$, $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ is not cyclic
In $G=(\Bbb Z/4\Bbb Z)^{\times}$, we have the Cayley table
$$\begin{array}{c|cc} \times_4 & 1+4\Bbb Z & 3+4\Bbb Z\\ \hline 1+4\Bbb Z & 1+4\Bbb Z & 3+4\Bbb Z\\ 3+4\Bbb Z & 3+4\Bbb Z & 1+4\Bbb Z \end{array}.$$
Thus $G$ is cyclic with generator $3+4\Bbb Z$.