Which of the following statements are true about finite cyclic groups? $1.$ If $G$ is a finite cyclic group, then for every subgroup $H,K$ of $G,$ we have either $H\subset K$ or $K\subset H.$
$2.$ If for every subgroup $H,K$ of a finite group $G,$ we have either $H\subset K$ or $K\subset H,$ then $G$ is cyclic.
I have figured a counterexample for the first statement:
Let $G=\mathbb{Z}_{6}.$ Then $H=2\mathbb{Z}_{6}$ and $K=3\mathbb{Z}_{6}$ are subgroups of $\mathbb{Z}_{6}$ but neither $H\subset K$ nor $K\subset H.$ So the first statement is wrong.
However, I am not sure about the 2nd statement.
 A: The statement is true. Suppose $G$ is not cyclic, and take an element $g\in G$ of maximal order. Such an element exists because the group is finite. By assumption $\langle g\rangle\ne G$, and so there is some element $x\in G\setminus\langle g\rangle$. Now let $H=\langle g\rangle$ and $K=\langle x\rangle$. We will show that none of these subgroups is contained in the other, and this will be a contradiction.
Since $x\in G\setminus\langle g\rangle=G\setminus H$ it follows that $x\notin H$, and so $K$ is not contained in $H$. So we just have to show $H$ is not contained in $K$. Let's assume $H$ is contained in $K$. Then there is some $n\in\mathbb{N}$ such that $g=x^n$. It is well known that if $ord(x)=m$ then $ord(g)=ord(x^n)=\frac{m}{\gcd(m,n)}$. If $\gcd(m,n)>1$ then it would mean that $ord(g)<ord(x)$, contradicting the assumption that $g$ has maximal order. So we must have $\gcd(m,n)=1$. Thus there are some $k,l\in\mathbb{Z}$ such that:
$$km+ln=1$$
But then note that $x=x^{km+ln}=(x^m)^k(x^n)^l=g^l$, and so $x\in\langle g\rangle$, a contradiction.
