Exercise 2.3.23 (If $n\geq 3$, $(\mathbb{Z}/2^n \mathbb{Z})^\times$ is not cyclic)

Proof. Since $n>1$, $1\not\equiv -1(\mathit{mod}{2}^n)$, so $-\bar{1}$ is an element of order $2$ in $G={(\mathbb{Z}∕2^n\mathbb{Z})}^{\text{×}}$. Thus

is a subgroup of $G$ of order $2$.

By Exercise 22, if $n\ge 3$, $\bar{5}$ has order $2^{n-2}$ in $G$. Therefore ${\bar{5}}^{2^{n-3}}$ has order $2$ in $G$. This gives a subgroup

of order $2$.

We show by contradiction that these two subgroups are distinct. Assume that $H=K$. Then ${\bar{5}}^{2^{n-3}}=\overline{-1}$, so $5^{2^{n-3}}\equiv -1(\mathit{mod}{2}^n)$. But we know by Exercise 22 that $5^{2^{n-3}}\equiv 1+2^{n-1}(\mathit{mod}{2}^n)$. This gives $1\equiv -1+2^{n-1}(\mathit{mod}{2}^n)$, so $2^{n-1}\mid 2$. Since $n\ge 2$, $2\mid 2^{n-1}$, thus $2=2^{n-1}$, so $n=2$. This contradicts the hypothesis $n\ge 3$. Therefore $H\ne K$.

By Theorem 7, part 3, every cyclic group of even order has exactly one subgroup of order $2$. Since $|G|$ is even and $G$ has two distinct subgroups of order $2$, $G$ is not cyclic.

In conclusion, ${(\mathbb{Z}∕2^n\mathbb{Z})}^{\text{×}}$ is not cyclic for any $n\ge 3$. □