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Consider the group $G = (\mathbb{Z}/m \mathbb{Z}, +)$. The statement I want to prove is that the cyclic subgroup generated by $[a]_m \in \mathbb{Z}/m \mathbb{Z}$ is $\mathbb{Z}/m \mathbb{Z}$ if and only if $\gcd(a,m) = 1$.

I first worked out a few examples and verified it seemed to hold, but am struggling to get any intuition as to why, or how to start the proof. I know that $\gcd(a, m) = 1 \Leftrightarrow$ $[a]_m$ is invertible, but since we are in a group $G$ with operation $+$, everything is invertible regardless of its $\gcd$.

Shaun
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    Think about the order of $[a]_m$ in $G$. – Robert Shore Mar 26 '22 at 16:09
  • @RobertShore The order of $[a]_m$ will be infinite if $\gcd(a, m) = 1$. If $\gcd(a, m) > 1$, then the order would be finite for some $k$, and the cyclic subgroup would be ${a^0, a^1, \ldots, a^{k-1}}$, i.e, multiples of $a$ up to $(k-1) \cdot a$. But then the cyclic subgroup cannot be equal to $G$ because of those prime $i \leq m - 1$: they are not multiples of $a$, so they would not be in the cyclic subgroup? –  Mar 26 '22 at 16:17
  • That's not correct. Remember, this is an additive group, not a multiplicative group. What is $m[a]_m$ in $G$? – Robert Shore Mar 27 '22 at 08:21

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