# A proposition on cyclic group

$G$ is a cyclic group iff $$\forall H < G, \ \exists k, \ H = \{a^k : a \in G\}.$$

Is it right?

• Would you care to give some more explanation for how you came to this question, why you would like to know the answer, what you do and don't know about infinite group theory, and so on? Dec 16, 2014 at 4:38
• To Yemon Choi, it just a sudden idea. By that necessity is obviously correct I come to ponder over sufficiency. I've made a "proof" for $G$ is finite, but I haven't had a review. (The idea is considering about the element with the largest order) Dec 16, 2014 at 6:08

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.