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

Posted: February 12, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

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

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