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I'm trying to find an element $k$ that generates the cyclic additive group $\mathbb{Z}_{6}$. Since a group is cyclic, the entire group can be generated by a single element. I've tried adding $\langle1\rangle$ and $\langle5\rangle$ repeatedly in modulo $6$. And both $1$ and $5$ give me all the elements of $\mathbb{Z}_{6}$. So is it possible for a cyclic additive group to have more than one generator or am I doing something wrong?

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Yes, it is possible. No, you are not doing anything wrong. For example, every nonzero element generates $\mathbb{Z}_p$ when $p$ is prime. –  Qiaochu Yuan Jul 30 '11 at 3:01
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In fact, the cyclic group of order $n$ will in general have $\varphi (n)$ generators, where $\varphi$ is the Euler totient function. The fact that $\varphi (6) = 2$ has the interesting consequence that it is impossible to fold a strip of paper into a hexagonal knot... –  Zhen Lin Jul 30 '11 at 3:07
    
I know of a solitary (=patience) card game played with two decks that depends on the fact described in Qiaochu's comment in the special case $p=13$. According to the legend that game was invented (ages ago) by an Oxbridge math senior. –  Jyrki Lahtonen Jul 30 '11 at 9:11

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up vote 3 down vote accepted

You're absolutely right. More generally, if you have an element $a$ that generates a finite cyclic group $G$, the group is also generated by another element $a^n$ (in multiplicative notation, $n\cdot a$ in additive notation) iff $n$ is coprime to $|G|$, i.e. $\gcd(n,|G|)=1$. This is because in that case, you can write any exponent $k$ in $a^k$ as $k=rn+s|G|$, and thus

$$a^k=a^{rn+s|G|}=\left(a^n\right)^r+\left(a^s\right)^{|G|}=\left(a^n\right)^r\;.$$

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