# Is a group cyclic with its generators

A group (S, $\odot$) is called cyclic if there exists g $\in$ S such that for every a $\in$ S there exists an integer n such that a = g $\odot$ g ... $\odot$ g (n times). If such a g $\in$ S exists, it is called a generator.

Is the group $\mathbb{Z}^{*}_{13}$= {1, 2 ... 11, 12} together with multiplication modulo 13 then cyclic? I can't seem to find the generators

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You really didn't need to define a group, or even define cyclic group. –  Thomas Andrews Nov 19 '13 at 21:27
A group is cyclic if and only if it is generated by a single element. If there is no single element which generates the group, it is not cyclic. Can you find an element that generates $\mathbb{Z}_{13}^*$? –  IBWiglin Nov 19 '13 at 21:28
Allright, i'll edit it! –  user108287 Nov 19 '13 at 21:28

Without some background in number theory, all you can do is try various numbers. Obviously $1$ doesn't work. So try $2$. We get lucky, $2$ works. For let us find the various positive powers of $2$, reduced modulo $13$. We get, in order, $$2,4,8,3,6,12,11,9,5,10,7,1.$$

Remark: There are other generators, a total of $4$ of them. It turns out that they are $2^1$, $2^5$, $2^7$, and $2^{11}$ (modulo $13$).

At this early stage, if you want all the generators, it is best to compute. Let us test $3$. The various powers, modulo $13$, are $3$, $9$, $1$, and now things start all over again, so we certainly won't get everything.

Next let's try $4$. The powers of $4$ will be the even powers of $2$, so we can look back on the work we did with $2$ and see that we will only get $4$, $3$, $12$, $9$, $10$, and $1$.

Quite a few left. Let's try $5$. It turns out that $5$ is no good, because $5^4$ gives $1$. Continue.

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Thanks for the explanation. Isn't it so that in fact, if p is prime, the generators of Z$_p$ are 1, 2, ..., $p - 1$? So that would mean all the numbers from 1 to 12 are generators. –  user108287 Nov 19 '13 at 21:50
No. If $p$ is prime, and you look at the numbers $0$ to $p-1$, under addition modulo $p$, then $1,2,\dots,p-1$ are indeed all generators. However, your group is a different one, different number of elements, $12$ instead of $13$, and the operation is multiplication modulo $p$. –  André Nicolas Nov 19 '13 at 21:55
Thanks for the great explanation! –  user108287 Nov 19 '13 at 21:56
You are welcome. Concrete computational experience is very important, else this stuff may remain a bunch of mysterious "rules." –  André Nicolas Nov 19 '13 at 21:58

Try to determine the order of each element. Iff there is no element of order 12, the group isn't cyclic.

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It is cyclic. You can take for example $2$ as generator: $$1=2^{11},2=2^1,3=2^4,4=2^2,5=2^8,6=2^5$$ $$7=2^{10},8=2^3,9=2^7,10=2^9,11,12=2^6$$ In fact $\mathbb{Z}_p^*$ is prime for every prime $p$.

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