**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.23**

Show that $(\mathbb{Z}/(2^n))^\times$ is not cyclic for any $n \geq 3$. (Hint: find two distinct subgroups of order 2.)

Solution: Note that $(2^n-1)^2 = (-1)^2 = 1 \pmod {2^n}$, so $\langle 2^n-1 \rangle$ is a subgroup of order 2 in $(\mathbb{Z}/(2^n))^\times$. Moreover, by Exercise 2.3.22, $$(1 + 2^{n-1})^2 = 1 \pmod {2^n},$$ so $\langle 1 + 2^{n-1} \rangle$ is also a subgroup of order 2.

Clearly these two are not equal since $n \neq 1$; thus $(\mathbb{Z}/(2^n))^\times$ has two distinct subgroups of order 2. Thus $(\mathbb{Z}/(2^n)^\times$ is not cyclic.