Finite Group whose Subsets form Chain is Cyclic P-Group

Theorem

Let $G$ be a group.

Let $G$ be such that its subgroups form a chain.

Then $G$ is a cyclic $p$-group.

Suppose $G$ is not a $p$-group.

Then there exist two distinct primes $p_1, p_2$.

By Cauchy's Group Theorem, there exist subgroups $H, K$ such that:

$\order H = p_1$

$\order K = p_2$

That is:

$H = \gen a$

$k = \gen b$

for some $a, b \in G: a \ne b$, where:

$\order a = p_1$

$\order b = p_2$

and so both $H \nsubseteq K$ and $K \nsubseteq H$

Thus, by definition, the subgroups of $G$ do not form a chain.

It follows by the Rule of Transposition that $G$ is a $p$-group.

It remains to be shown that $G$ is cyclic.

Suppose to the contrary that $G$ is not cyclic.

Then it is not the case that $G$ is generated by a single element of $G$.

Thus there exist $a, b \in G, a \ne b$ such that $H = \gen a$ and $K = \gen b$ are subgroups of $G$.

But as $a \in H, b \in K, a \notin K, b \notin H$ it follows that $H \nsubseteq K$ and $K \nsubseteq H$.

It follows by the Rule of Transposition that $G$ is cyclic.

$\blacksquare$

1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 43 \beta$