Ytukyg

Group $G$ is cyclic $iff $

Every subgroup of $G$ is like $G^m = {g^m | gin G}$ for some $m in mathbb N$.

$G^m = {g^m | gin G}$ means $forall x in G^m, exists, g in G$ s.t. $x=g^m$.

$Rightarrow$ is trivial.$,$What about $Leftarrow$?

My idea:$,$If $G$ is infinite group, it's not very hard.

Suppose $a in G$ has infinite order, $langle arangle=G^m$ for some $m>0$, then $exists , b in G$ s.t. $a=b^m$. Then $langle brangle=G^n$ for some $n>0$, then $exists , c in G$ s.t. $b=c^n$. Since $c^m in G^m = langle arangle$, $exists ,i in mathbb N $ s.t. $c^m=a^i=b^{mi}=c^{nmi}$. $c$ clearly has infinite order, so $m=nmi$, $1=ni,n=1$.$langle brangle=G$ is cyclic.

So what if $G$ is finite?

Thanks for your time and help :)

If $G^m$ forms a group, then it is easy to find that $G^m$ is a characteristic subgroup of $G$.

Suppose every subgroup of $G$ is like $G^m={g^m | gin G}$ for some $min mathbb{N}$, then every subgroup of $G$ is a characteristic subgroup. It follows that $G$ is abelian.

Suppose $G$ is finite. $Gcong mathbb{Z}_{n_1}timescdotstimesmathbb{Z}_{n_s}$ for some positive integers such that $n_1|n_2|cdots|n_s$. Pick $H = mathbb{Z}_{n_1}cong G^m$ as a subgroup of $G$ for some $minmathbb{N}$ and then the quotient group $G/Hcongmathbb{Z_2}timescdotstimesmathbb{Z}_{n_s}$. Note that the exponent of $G/H$ is $n_s$, we have $n_s|m$. However, the exponent of $G$ is also $n_s$, which implies that $g^{n_s} = 1$ and so $g^m = 1$ for all $gin G$. It follows that $G^m = {1}$. This makes $s = 1$ and thus $G$ is cyclic.