Generator of Additive Group Modulo m iff Unit of Ring
Theorem
Let $m \in \Z: m > 1$.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Let $a \in \Z_m$.
Then:
- $a$ is a generator of $\struct {\Z_m, +_m}$
- $a$ is a unit of $\struct {\Z_m, +_m, \times_m}$
Proof
From Integers under Addition form Infinite Cyclic Group, the identity element $1_\Z$ of the ring of integers $\struct {\Z, +, \times}$ is a generator of the group $\struct {\Z, +}$.
Thus from Quotient Group of Cyclic Group, the identity element $1_{\Z_m}$ of the ring $\struct {\Z_m, +_m, \times_m}$ is a generator of the group $\struct {\Z_m, +_m}$.
Let $a \in \Z_m$.
Suppose $1_{\Z_m} \in \gen a$, where $\gen a$ signifies the group generated by $a$.
Then the smallest subgroup of $\struct {\Z_m, +_m}$ containing $1_{\Z_m}$, that is $\struct {\Z_m, +_m}$ itself, is contained in $\gen a$.
Thus $\gen a = \struct {\Z_m, +_m}$ iff $1_{\Z_m} \in \gen a$.
However, from Subgroup of Additive Group Modulo m is Ideal of Ring, $\gen a$ is an ideal of $\struct {\Z_m, +_m, \times_m}$, and hence is the principal ideal $\ideal a$ generated by $a$.
But from Principal Ideal from Element in Center of Ring, $1_{\Z_m} \in \gen a$ if and only if $a$ is a unit of the ring $\struct {\Z_m, +_m, \times_m}$.
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.9$