$G$ is group , $|G|=n$ , suppose every $d$ divides $n$ , there are $d$ elements so $x^d=1$ , prove $G$ is cyclic.
we have prime factorization of $n=$$p_1^x....p_k^z$.
define $$H=p_j^i$$ because its a p-group there is T subgroup of H , order |$(p_j)^(i-1))$| .
take $g ∈ H , g∉T , o(g)=$$p_j^i$ because if o(g)=$p_j^(i-1)$ there are too much elements exist $(g)^(p_j)^(i-1) = 1$
i dont know how to generalize it for every G and not just for p-group.
Very quickly, if you want to continue your line of reasoning: once you have the (unique) Sylow $p$ of $G$ (for every prime $p$) is cyclic, apply
Theorem: If $H,K<G$ normalises one another, then $[H,K]\unlhd HK$ and $[H,K]\leq H\cap K$. In particular, if $H\cap K$ is trivial then $HK\cong H\times K$.
Thus $G$ is the direct product of its Sylows (clearly with pairwise coprime orders) and so $G$ is cyclic.
Let $G_d =\{g \in G : o(g) = d\}$ be set of elements in $G$ whose orders are exactly $d$. Then, since order of element in $G$ is a divisor of $n$, $\sum_{d|n}|G_d| = n$.
Claim : $|G_d| = \varphi(d)$, where $\varphi$ is Euler totient function.
(proof) Note that $|G_1|=1$. Let $p \mid n$, then since there are $p$ elements, $1_G$ and $(p-1)$ number of nonunit elements, which satisfy $x^p =1$, $|G_p| = p-1= \varphi(p)$; similarly, $|G_{p^k}| = p^{k-1}(p-1)= \varphi(p^k)$. Note that $d = |\{x\in G : x^d = 1\}| = \sum_{k\mid d }|G_d|$, and since $d = \sum_{k\mid d} \varphi(k)$, inductively we have $|G_d| = \varphi(d)$. (end of the proof)
Thus, since $|G_n| = \varphi(n) \ge1$, there exist an element $x$ such that $o(x) = n$. $G = \{1, x, x^2, \dots, x^{n-1}\}\stackrel{\sim}{=}\mathbb{Z}/n\mathbb{Z}$ is cyclic.
