Is there a noncyclic group of order 39?

[Bufo]

The answer is no. But I'm supposed to solve this using basics of isomorphism (without any application of Sylow's theorem).

Suppose such a group exists. Then of course it's non-Abelian.

I know there are cyclic subgroups of order 3 or 13 (actually both, because possibly I'm allowed to use this particular implication of Sylow's theorem).

 

[ygolo]

If you're allowed to use that little tid-bit (seems redundant since LaGrange's theorem will give you that it will have one such cyclic supgroup if the whole group itself is not cyclic), then you are essentially done, because Z_3xZ_13 is isomorphic to Z_39. <(1,1)> generates the whole group.

You still have to prove that since every group of order 39 has sub-groups of order 3 and of order 13, that it has to be isomorphic to Z_3xZ_13.

It's been a while since I took group theory, but I think you can go about it by proving one or the other fo the supgroup of 3 or 13 is Normal in your original group. Then the factor group must be cyclic (again using LaGanges Theorem).

That's just the sketch of a proof that popped into my head. But like I said, it has been a long time since my Math degree.