If you find any mistakes, please make a comment! Thank you.

The group of units in $\mathbb Z/(2^n)$ is not cyclic for n at least 3


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.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu