# Generators of a finite additive cyclic group

Let $C$ be an abelian additive group and write e for a generator of $C$. The elements of $C$ are then $0,e,2e,3e,\dots,(n-1)e$. If $C$ is finite, prove that the element $ke$ is another generator of $C$ if and only if $k$ and $n$ are relatively prime.

• Can you argue the 'if' and/or the 'only if' part? – Berci Nov 12 '12 at 15:24
• I do not believe so. – user47636 Nov 12 '12 at 15:25

Suppose gcd$(n, k) = 1$. Supose $mke = 0$ for an integer $m$. Then $n|mk$. Since gcd$(n, k) = 1$, $n|m$. Hence the order of $ke$ is $n$.

Conversely suppose $d =$ gcd$(n, k) \ne 1$. Let $k' = \frac{k}{d}, n' = \frac{n}{d}$. Then $n'ke = n'dk'e = nk'e = 0$. Since $n' < n, ke$ is not a generator.

• Can you elaborate why from $mke = 0$, $n | mk$ follows? – black Dec 24 '18 at 13:52

Here is one direction: Let $\gcd(k,n) = 1$ then there exist $i, j \in \mathbb Z$ such that $1 = ik + jn$. If $me$ is an arbitrary element in $C$, $m \in \mathbb Z$, then $me = m(ik + in) e = mik e + min e = mike = (mi)ke$.

The statement is a actually a direct consequence of the following theorem:

Let $a$ be an element of order $n$ and let $k$ be a positive integer. Then $| a^k| = n / \gcd(n,k)$.

• See for example Gallian, 7th Edition, page 75, Theorem 4.2. – Rudy the Reindeer Nov 12 '12 at 15:39