Prove or disprove that there is an abelian, noncyclic group of order 52.

Question

So I've heard one must invoke Sylow's theorems in order to break down something like this. So far I know that there is a subgroup of order 13 in G, and that it's the only subgroup of order 13 in G. To go from there to proving that it's abelian or cyclic or not seems trickier. Helpful hints appreciated, thank you.

Answer 1

Hint: Consider a direct product of cyclic groups whose orders multiply to $52$, but such that the product is not cyclic. The fact that

$$52 = 2^2 \cdot 13$$

is highly relevant here.

Answer 2

Forget Sylow theorem. Note $52=2.26$ and consider $$\mathbb Z_2\oplus\mathbb Z_{26}$$

Of course this one is abelian (the components being so). Is it possible to find $(a,b)\in\mathbb Z_2\oplus\mathbb Z_{26}$ such that $l.c.m.\{|a|,|b|\}=52?$

Answer 3

Let G be any group of order 52. Then by sylow theorem it has only one sylow 13 subgroup and hence is normal in G, let it be H. Then G/H forms a quotient group of order 5, which implies G/H is cyclic and hence G is abelian.