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