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?
A very basic proof:
If $G$ has an element of order $5$ and an element of order $7$, say $a$ and $b$, then since $G$ is abelian, $ab$ has order $35$ and $G$ is cyclic.
So we want to rule out the possibility that $G$ has no elements of order $5$, or alternatively, no elements of order $7$, without recourse to Cauchy's Theorem.
Elements of order $5$ occur "$4$ at a time" (any element of order $5$ generates a subgroup of order $5$, and any two such subgroups intersect in just the identity, so that gives $4$ elements of order $5$ in each distinct subgroup of order $5$), and similarly, elements of order $7$ occur "$6$ at a time".
Now $G$ has $34$ non-identity elements, which is neither a multiple of $4$, nor $6$, so we cannot have elements of only one order among the non-identity elements.
P.S.: the condition that $G$ be abelian is unnecessary-any group of order $35$ is actually cyclic, but a proof of this will have to wait until you have access to more advanced theorems.
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.
