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…
Best Answer
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 $\prod_{i=1}^ke_i=n$.