Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
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
add comment

1 Answer 1

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


share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.