Question

If all proper nontrivial subgroups of a nontrivial group are isomorphic to each other, must G be cyclic?

Answer 1

Suppose that G has exactly two nontrivial proper subgroups H and K.

If K has a proper nontrivial subgroup, then H≤K. Thus K has exactly one proper nontrivial subgroup, which means that K is cyclic of order p2. Hence G must be a p-group of order p3, and any element x∉K will generate G.

If H and K do not have proper nontrivial subgroups, H and K will be cyclic of prime order. It is clear that H and K will be normal subgroups and that G is cyclic of order pq.

You try can classify groups with exactly n subgroups for small n. This is the case n=4. You can try to do the case n=5 by using the classification of groups with exactly n=1,2,3,4 subgroups. A starting point is to notice that if a group with n subgroups contains a subgroup with n−1 subgroups, it will be cyclic.

After that you can think about how the number of subgroups affects the structure of groups. What is the largest k such that every group with ≤k subgroups is cyclic

There are two a priori possibilities:

(1) It could be that G has no nontrivial subgroups; then the condition "G is isomorphic to each one of its nontrivial subgroups" is true by vacuity. This can only happen if G itself is trivial (if G≠{1}, let x∈G, x≠1; then 〈x〉 is a nontrivial subgroup of G). So we could have G={1}, not contemplated in the original question.

(2) It could be that G≠{1}. Then G must be cyclic: if x∈G, x≠1, then 〈x〉 is a nontrivial subgroup of G, hence G≅〈x〉 is cyclic. Thus, either G≅Z, or G≅Cn for some n>1.

If G≅Z, then G is isomorphic to any nontrivial subgroup.

If G≅Cn, and d|n, then G has a subgroup of order d which is cyclic. If d>1, then G≅Cd≅Cn, so d=n. Thus, the only divisors of n are 1 and n, hence n is prime. We change its name to p, because that is the only way in which the original statement will be true, and we conclude that G≅Cp, as desired