$G$ is a cyclic group iff $$ \forall H < G, \ \exists k, \ H = \{a^k : a \in G\}. $$
Is it right?
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Sign up to join this community$G$ is a cyclic group iff $$ \forall H < G, \ \exists k, \ H = \{a^k : a \in G\}. $$
Is it right?
A bit of searching revealed the following reference for this statement in question
F. Szasz, On cyclic groups, Fund. Math., 43(1956), 238-240
In the following more recent paper the authors proved a refinement. Let $k$ denote the number of subgroups of $G$ that are not of the form $\langle a^n,a\in G\rangle$. The main theorem says that $k=0$ iff $G$ is cyclic, $1\le k<\infty$ iff $G$ is finite non-cyclic, and $k=\infty$ iff $G$ is infinite non-cyclic.
W. Zhoua, W. Shib, Z. Duan, A new criterion for finite non-cyclic groups Communications in Algebra, 34 (2006), 4453-4457
$\def\ord{\mathop{\rm ord}}\def\dvds{\mathrel{|}}$Yes, this is true. The argument is similar to your `maximal order' argument; this shows again that it is helpful to provide your thoughts in the question.
For $H<G$, denote by $f(H)$ the minimal $k$ such that $H=\{a^k\colon a\in G\}$. We notice that if $H_1<H_2<G$ then $k_1=f(H_1)$ is divisible by $k_2=f(H_2)$. Indeed, for every $a\in G$ we have $a^{k_1},a^{k_2}\in H_2$, so $a^{\gcd(k_1,k_2)}\in H_2$, and $H_2\supseteq \{a^{\gcd(k_1,k_2)}\colon a\in G\}$. The converse inclusion is obvious, thus $H_2=\{a^{\gcd(k_1,k_2)}\colon a\in G\}$ and hence $\gcd(k_1,k_2)\geq k_2$, as required.
Now, we have an alternative: either (i) there exists an infinite chain $H_1<H_2<H_3<\dots<G$ or (ii) each cyclic subgroup is contained in a maximal cyclic subgtoup. The case (i) is ruled out since we should have then $\dots \dvds f(H_3)\dvds f(H_2)\dvds f(H_1)$ which is impossible.
In case (ii), consider a maximal syslic subgroup $H=\langle a\rangle$. If $H=G$ we are done. Otherwise $k=f(H)>1$, and $a=b^k$ for some $b\in G$. If $b\notin H$ then $\langle b\rangle > H$, which contradicts maximality. Thus $b=a^\ell$; this means that $\ord a$ is finite (denote $n=\ord a$) and $\gcd(n,k)=1$. For every $g\in G$, we have $g^{nk}=e$. Thus we see that the orders of elements are bounded, and we may assume that $\ord a$ is maximal (then $\langle a\rangle$ is still a maximal cyclic subgroup).
If for every $g\in G$ we had $g^n=e$ (and $g^k\in H$) then we would have $g=g^{\gcd(n,k)}\in H$, which would imply $H=G$. Thus there exists some $g\in G$ with $g^n\neq e$ (and $\ord g\dvds nk$). Then $1<\ord g^n\dvds k$. Both cyclic subgroups $H$ and $\langle g^n\rangle$ are normal (due to the initial condition), and their orders are coprime, so they form a direct product, which contains an element of order $n\cdot \ord g^n>n$. A contradiction.