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...