Prove that a group G has exactly 3 subgroups if and only if G is cyclic with |G|

ID: 3028699 • Letter: P

Question

Prove that a group G has exactly 3 subgroups if and only if G is cyclic with |G| = p^2 for a prime number p.

Explanation / Answer

1) Consider such a group G. Suppose there are elements a and b in this group of prime orders p and q, respectively. Then the cyclic groups generated by a and b are two different proper subgroups of G, a contradiction. Thus every element of G must have order a power of a single prime number p. Pick an element a in G. If |a| = p^k with k > 2, then a^p and a^(p^2) generate two different proper subgroups of G, a contradiction. Thus every element must have order either p or p^2. Suppose every element has order p. Then take a in G. If a generates G, then G will be a cyclic group of order p that has no proper subgroups, a contradiction. Thus |G| > p. Then there is an element b in G but not in the cyclic group generated by a. The cyclic subgroup of G generated by b has order p, too, and is distinct from that generated by a. This gives us two different proper subgroups of G, a contradiction. Thus G must contain an element a of order p^2.

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Prove that a group G has exactly 3 subgroups if and only if G is cyclic with |G|

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