## Finite groups with at most one subgroup of any order are cyclic

Posted: February 12, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Obviously every finite cyclic group has at most one subgroup of any order. We are going to prove the converse of this fact, i.e. if a finite group $G$ has at most one subgroup of any order, then $G$ is cyclic.

Lemma. Let $G$ be a finite $p$-group. If $G$ has at most one subgroup of any order, then $G$ is cyclic.

Proof. Let $a \in G$ be such that $o(a)=p^n$ is as large as possible and put $H = \langle a \rangle.$ Let $b \in G.$ Then $o(b)=p^m,$ for some $m \leq n.$ Thus $p^m \mid p^n.$ We know that a $p$-group has a subgroup of any order which divides the order of the group. Thus $H$ has a subgroup $K$ of order $p^m.$ But $\langle b \rangle$ has also order $p^m$ and so $K = \langle b \rangle$ because, by our hypothesis, $G,$ and therefore $H,$ can have at most one subgroup of any order. So $\langle b \rangle \subseteq H$ and hence $b \in H.$ We have proved that every element of $G$ is in $H$ and so $G=H. \ \Box$

Problem. Let $G$ be a finite group. If $G$ has at most one subgroup of any order, then $G$ is cyclic.

Solution. Let $\{P_1, \cdots , P_n \}$ be the set of all Sylow subgroups of $G$ and let $g \in G.$ Clearly $|P_i|=|gP_ig^{-1}|,$ for all $i,$ and so $P_i=gP_ig^{-1}$ because $G$ has at most one subgroup of any order. So $P_i$ are normal in $G$ and hence

$G = P_1 \times P_2 \times \cdots \times P_n. \ \ \ \ \ \ \ \ \ \ \ (*)$

On the other hand, $P_i$ cannot have more than one subgroup of any order because every subgroup of $P_i$ is also a subgroup of $G.$ Thus, by the above lemma, all $P_i$ are cyclic. The result now follows from $(*)$ and the fact that $\gcd(|P_i|, |P_j|)=1$ for all $i \neq j. \ \Box$