Prove that $G$ must be cyclic 
Let $G$ be a finite group, assume that for any pair of subgroup $H, K$ of $G$ either $H\subseteq K$ or $K\subseteq H$. Prove that $G$ must be cyclic. 


Suppose $a$ is maximum in $G$. Let $\langle a\rangle =H$ and $\langle b\rangle =K$. If $H\subseteq K$, then it is a contraction since $a$ is maximum. If $K\subseteq H$, then $b=a^n\in H$ where $n\in \mathbb{N}$, hence $H$ is cyclic which implies that $G$ is cyclic.

Can anyone check the proof valid or not? thanks
 A: I think you have the right idea and are simply expressing it in an unclear way. To use basically the same structure as you are and the same proof method, one might write it as:

Let $a$ be such that $\langle a \rangle$ is never strictly contained in $\langle b \rangle$ for any $b$. If we let $H=\langle a \rangle$ and $K=\langle b\rangle$ for some $b$, then we have that either $H\subseteq K$ or $K\subseteq H$. However, given that we chose $a$ such that $\langle a\rangle \not\subset \langle b \rangle$ for any $b$, we can assume that $K\subseteq H$, implying that $a^n=b$. Given that this holds of any $b$, we conclude that $a$ must generate the whole group, meaning the group is cyclic.

We might also consider justifying the first sentence (e.g. why such an $a$ exists). We can consider $a$ a maximal element in $G$ when ordered by the relation $a>b$ where $\langle a\rangle \supset \langle b\rangle$ - and every finite poset has to have a maximal element, meaning such an $a$ exists.
A: Perhaps one rewrite the proof along the following lines.
Let $a$ be an element of $G$ of maximal order, and let $H$ be the subgroup of $G$ generated by $a$.  We show that $H$ is all of $G$.
Suppose to the contrary that there is a $b\in G$ such that $b$ is not in $H$, and let $K$ be the subgroup of $G$ generated by $b$. It is clear that $K$ is not a subset of $H$. So $H$ is a subset of $K$, necessarily proper. This contradicts the maximality of the order of $a$.
A: Suppose $G$ is not cyclic, so let $\{a_1,\dots,a_n\}$ be the set of generators and let $H_i=\langle a_i \rangle$ where $i\in \{1, \dots n\}$. 
Now by hypothesis, we will get $$H_{i_1}\subseteq H_{i_2} \dots \subseteq H_{i_n}=\langle a_{i_n}\rangle$$ and this implies that $a_{i_n}$ generates $G$ and thus cyclic which is a contradiction. $\hspace{2cm} \blacksquare$
