TITLE: The group of invertible elements of $\mathbb F_{p}[x]/(x^m)$ is not a cyclic group.
QUESTION [8 upvotes]: I am stuck in a question about finite fields and would like to ask you for some help.

Given an integer $m\geq 2$ and $p$ a prime number, show that  $(\mathbb F_{p}[x]/(x^m))^{\times}$ (the group of all invertible elements of $\mathbb F_{p}[x]/(x^m)$) is not a cyclic group.

I don't really know how to start. I know a lemma which tells me that for every finite field $\mathbb F_{q}$ with $q$ elements, the multiplicative group of the nonzero elements $\mathbb F_{q}^{\times}$ is cyclic. But I don't think that I should use it here...
Does anyone have an idea? Thank you in advance!

REPLY [2 votes]: A cyclic group has at most one subgroup of each order, to it is enough to show that there exists two invertible elements in $F_p[x]/(x^m)$ which have the same order and generate different subgroups. In particular, in a cyclic group there is at most one element of order two.
If $p$ is odd, there is a nonzero element $\alpha$ in the field which is different to $1$, and the elements $a=1+x^{n-1}$ and $b=1+\alpha x^{n-1}$ are two different units of order two.
If $p$ is even and $m\geq4$, then $a=1+x^{m-2}+x^{m-1}$ and $b=1+x^{m-1}$ are two different elements of order two.
We are left with $p=2$ and $m=3$, which you can do by hand :-)