# Generator of cyclic groups

I have a group of order $13\times 11\times 7$. I am able to show that my group is abelian (using a combination of Sylow's theorems and seeing that the intersection of the 3 cyclic subgroups $C_{13},C_{11}, C_7$ is trivial.)

Now comes the deceptively simple part. I have to deduce from these information that my group is cyclic.

This is what I don't understand: Why do we have to show that the group is abelian? Doesn't the fact about the trivial intersection imply that we have an element of order $13\times 11\times 7$ which we can form by taking $g_{13}\circ g_{11}\circ g_{7}$? where $g_k$ are the generators of each cyclic group? How can I form a rigorous argument? I am pretty sure that the fact that my group is abelian does not directly imply that it is cyclic? Since there are finite abelian groups that are non-cyclic...

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What is the order of $g_{13}\circ g_{11}\circ g_{7}$? –  N. S. Jan 26 '12 at 13:46
In an abelian group, if $a$ has order $m$ and $b$ has order $n$, what is the order of $ab$? –  lhf Jan 26 '12 at 13:49
So, what group exactly are you dealing with? $\mathbb{Z}_{13}\times\mathbb{Z}_{11}\times\mathbb{Z}_{7}$ ? –  Thomas E. Jan 26 '12 at 13:55
@Gerrard You want to say that this element $g_{13}g_{11}g_7$ has order $13 \cdot 11 \cdot 7$. Let's put ourselves in a simpler situation: say I have elements $x, y$ in a group of prime orders $p, q$ where $p \neq q$, so that the cyclic subgroups they generate intersect trivially. Does $xy$ have order $pq$? Actually, this breaks down for the first nonabelian group that one thinks of: in $S_3$, $[12][123] = [23]$ has order $2$. –  Dylan Moreland Jan 26 '12 at 14:32
@N.S.: Oh!! is the answer to the last question be because the group is abelian therefore we can write $(g_{13}g_{11}g_7)^n$ as $g_{13}^ng_{11}^ng_7^n$. But if the group were not abelian this argument won't apply and we can have $(g_{13}g_{11}g_7)$ be of some other order not equal to 13 times 11 times 7? –  Gerrard A Jan 26 '12 at 15:21

A group $G$ may be generated by two elements $a$ and $b$ of coprime order and yet not be cyclic.

The simplest family of examples is that of the dihedral groups $D_n$ with $n$ odd. The group $D_n$ is defined to be the group of plane isometries sending a regular $n$-gon to itself and it is generated by the rotation of $2\pi/n$ radians and any axial symmetry. These two isometries have orders $n$ and $2$ respectively, yet they don't commute.

Instead, if $G$ is abelian and generated by two elements of coprime order, then $G$ is cyclic. This can be done in two steps:

(1) prove that if $G=\langle a,b\rangle$ with ${\rm ord}(a)=m$, ${\rm ord}(b)=n$ and ${\rm GCD}(m,n)=1$, then $G\simeq\Bbb Z/m\Bbb Z\times\Bbb Z/n\Bbb Z$;

(2) prove that $\Bbb Z/m\Bbb Z\times\Bbb Z/n\Bbb Z\simeq\Bbb Z/mn\Bbb Z$ if ${\rm GCD}(m,n)=1$.

Maybe it's worth reminding the structure theorem for finite abelian groups: A finite abelian group $G$ with $|G|=n$ is always isomorphic to a product $C_1\times C_2\times\cdots\times C_k$of cyclic groups with $|C_i|=e_i$,such that $e_1\mid e_2\mid\cdots\mid e_k$ and $\prod_{i=1}^ke_i=n$.

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You don't have to first probe it's abelian; that will fall out. By Sylow's theorems, the subgroups of order 7, 11, and 13 are unique. Use this fact to show any subgroups of order 77, 91 and 143 are cyclic and also unique. (Hint: a subgroup of order 77 contains the unique subgroups of order 7 and 11 and these only account for 17 of the 77 elements; all the rest have order 77.) 1, 7, 11, 13, 77, 91 and 143 are the only proper factors of 1001, hence the only possible orders of proper subgroups. How many elements have been accounted for?

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