Let $G$ be a group (finite or infinite).

Given $k \in \Bbb N_+$, define $G^k=\{g^k\mid g\in G\}$

I hope to prove that $G$ is cyclic if and only if every subgroup of $G$ has the form of $G^k$.

It's easy to see that if $G=\langle g\rangle$ is cyclic, then every subgroup $H$ of $G$ must have the form of $\langle g^k \rangle$. Therefore $H=\langle g^k \rangle=\langle g \rangle^k=G^k$.

...excluding the trivial subgroup {e} of an infinite group (complement)

But I get into trouble proving the other side.

I have no idea how to find a counterexample, i.e. there exists a subgroup of $G$ not having the form of $G^k$ if $G$ is not cyclic.

Meanwhile, I tried to prove it directly. But I cannot even prove $G$ is Abelian, considering $G^2$ as a normal(proved) subgroup of $G$.

Could you please help me solve the problem or disprove this proposition?

Thanks a lot! And sorry for my poor English..

  • $\begingroup$ Can you prove that in any group $G$ (and therefore specifically in yours), every subgroup of the form $G^k$ is normal? $\endgroup$
    – Arthur
    Oct 9, 2017 at 5:50
  • $\begingroup$ Sure. $g^{-1}a^kg=(g^{-1}ag)^k\in G^k\implies g^{-1}G^kg=G^k$. $\endgroup$
    – kellty
    Oct 9, 2017 at 5:53
  • $\begingroup$ Cool. I don't know whether that would actually help, but it seems like it should... $\endgroup$
    – Arthur
    Oct 9, 2017 at 5:54
  • $\begingroup$ I hope so.But I cannot continue the proof.. $\endgroup$
    – kellty
    Oct 9, 2017 at 5:55
  • 1
    $\begingroup$ Sorry but in $S_3$ every subgroup has the form $G^k$ but still not cyclic.Is the problem correct? $\endgroup$
    – Learnmore
    Oct 9, 2017 at 6:01

2 Answers 2


Anyway, I'd appreciate for everyone's help. I have got the proof by myself (aha!), as following:

For the trivial subgroup $\{e\}$, there exists some positive integer $n$ so that $\{e\}=G^n$. Then

$\forall a\in G, a^n=e$, which means $o(a)$ (the order of $a$) is a factor of $n$.

As $A=\{m\in\mathbb{N^*}\mid\exists a\in G,o(a)=m\}$ is bounded (namely, $n$ is an upper bound), thus finite, we can get a finite set $B=\{p\in\mathbb{N^*}\mid \exists a\in G\text{ so that }p\text{ is a prime factor of }o(a)\}=\{p_1,\cdots,p_r\}$.

Next, every subgroup $H=G^k$ of $G$ is normal. Notice that

$\forall g\in G,\forall a^k\in H,ga^kg^{-1}=(gag^{-1})^k\in H\implies gHg^{-1}=H$.

According to Sylow's first theorem, $\forall p_i\in B,\exists g_i\in G,s.t. o(g_i)=p_i$. I'd show that $g_i$ is in the center of $G$.

$\forall a\in G,ag_ia^{-1}\in\langle g_i\rangle\implies\exists t\in\mathbb{N},s.t.ag_ia^{-1}=g_i^t$ (if $p_i|t$, then $g_i=a^{-1}ea=e$.Thus without loss of generality, $0<t<p_i$)

Meanwhile, $g_i^{-1}ag_i\in\langle a\rangle\implies\exists s\in\mathbb{N},s.t. g_i^{-1}ag_i=a^s$

We can get $g_ia^s=ag_i=g_i^ta\implies a^{s-1}=g_i^{t-1}\in\langle g_i\rangle$.

For $\gcd(t-1,p_i)=1$, $\langle g_i^{t-1}\rangle=\langle g_i\rangle\implies g_i\in\langle g_i^{t-1}\rangle=\langle a^{s-1}\rangle\implies g_i\in\langle a\rangle\implies g_i$ commutes with $a$. As $a$ is arbitrary, $g_i \in Z(G)$.

Let $g_0=g_1\cdots g_r\in Z(G)$, it's obvious that $o(g_0)=p_1\cdots p_r$.

Then $C=\{a\in G\mid p_1\cdots p_r|o(a)\}\neq\varnothing$, $D=\{m\in\mathbb{N^*}\mid\exists a\in C,o(a)=m\}$ is a bounded non-empty subset of $\mathbb{N}$. We can get $M=\max D$ and select some $h\in C$ so that $o(h)=M$.

There exists some $l$ so that $\langle h\rangle=G^l$. Particularly, $\exists b\in G,s.t. h=b^l$, and $o(h)=o(b^l)=\dfrac{o(b)}{\gcd(l,o(b))}|o(b)$

On one hand, $o(h)\leqslant o(b)$; on the other hand, $b\in C\implies o(b)\leqslant M=o(h)$. Thus $o(h)=o(b)\implies\gcd(o(b),l)=1\implies\gcd(p_1\cdots p_r,l)=1$.

Then $\forall a\in G, \gcd(o(a),l)=1$ ($o(a)$ must have the form of $\prod_{j=1}^rp_j^{\alpha_j}$). $\exists u,v\in\mathbb{Z},s.t. uo(a)+vl=1\implies a=a^{uo(a)+vl}=(a^v)^l\in G^l=\langle h\rangle$.

Therefore we got the consequence, $G=\langle h\rangle$ is cyclic.

: )

  • $\begingroup$ I hope that no gap is left...and by the way, I'm sincerely looking forward to anyone's better proof. $\endgroup$
    – kellty
    Oct 9, 2017 at 16:00
  • 1
    $\begingroup$ I haven’t read everything yet, but something seems strange. You don’t exclude the case of a infinite cyclic group, aka $\mathbb{Z}$. But then there is no $k$ with ${e}=G^k$ if you want $k>0$. But you can of course just exclude this case. $\endgroup$
    – Maik Pickl
    Oct 9, 2017 at 16:15
  • $\begingroup$ @MaikPickl well..it's to show that $G$ is cyclic if $G$ satisfies such condition... $\endgroup$
    – kellty
    Oct 9, 2017 at 16:16
  • 1
    $\begingroup$ True, but then the other direction fails for the integers. And you wanted an if and only if. $\endgroup$
    – Maik Pickl
    Oct 9, 2017 at 16:17
  • 1
    $\begingroup$ @MaikPickl Thanks! I got it...the question has been corrected. $\endgroup$
    – kellty
    Oct 9, 2017 at 16:22

Consider $S_3$ ;

Every subgroup has the form $G^k$ but $S_3$ is still not cyclic.

The subgroups are $\langle (12)\rangle,\langle (23)\rangle,\langle (31)\rangle,\langle (123)\rangle$

  • $\begingroup$ not $(32)$ but $(13)$ $\endgroup$
    – kellty
    Oct 9, 2017 at 6:10
  • $\begingroup$ Wait...$S_3^{6k+2}=S_3^{6k+4}=\langle (123) \rangle$, $S_3^{6k+3}=\{e,(12),(23),(31)\}$,$S_3^{6k+5}=S_3^{6k+1}=S_3$,$S_3^{6k}=\{e\}$.. $\endgroup$
    – kellty
    Oct 9, 2017 at 6:23
  • $\begingroup$ $\langle (12) \rangle$ doesn't have the form.. $\endgroup$
    – kellty
    Oct 9, 2017 at 6:24
  • $\begingroup$ This doesn't work - which exponent gives a group of order $2$? The exponent $3$ gives elements of orders $1$ or $2$, but these do not form a subgroup. $\endgroup$ Oct 9, 2017 at 6:35

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