Abelian group and element order Suppose that $G$ is an Abelian group of order 35 and every element of $G$ satisfies the equation $x^{35} = e.$ Prove that G is cyclic.
I know that since very element of $G$ satisfies the equation $x^{35} = e$ so the order of elements are $1,5,7,35$. I know how to prove the situation when one element's order is 35; one is 5, one is 7; but I don't know how to prove when all elements' order is 5 or 7. Is this possible? If it is, how can $G$ be cyclic?

Something more general is true. Let $G$ be an abelian group whose order $n$ is square-free (that is no perfect square $\ne1$ divides that $n$. Then that group is cyclic.
Given that condition, we have a facorization into distinct primes: $n=p_1p_2\cdots p_k$.
Now for any prime dividing the order of a group, by Cauchy's theorem there is an element of that order. Take the elements $g_i$ of order $p_i$ for all $i$. Then the product of all these $g_i$'s is an element of order $n$, making the group cyclic.
