Let be an odd prime. Prove that if is a noncyclic -group then contains a normal subgroup with . Deduce that for odd primes a -group that contains a unique subgroup of order is cyclic. (Note: for , it is a theorem that the generalized quaternion groups are the only noncyclic 2-groups which contain a unique subgroup of order 2.)

Let be an odd prime and let be a -group.

We will proceed by induction on the *breadth* of ; that is, on the number such that .

For the base case, let and let be a noncyclic group of order . Then by Corollary 4.9. Then is a normal subgroup of isomorphic to .

For the inductive step, suppose that for some , every noncyclic group of order contains a normal subgroup isomorphic to . Let be a noncyclic group of order . Since is a -group, is nontrivial. By Cauchy’s Theorem, there exists an element of order , and is normal in . Now is a -group of order .

Suppose is cyclic. By the Third Isomorphism Theorem, we have . Since is cyclic, every quotient of is cyclic, so that is cyclic. By this previous exercise, is abelian. Now is a finite abelian group of rank at least 2, since is not cyclic. Let be the -power map. By this previous exercise, is the elementary abelian -group whose rank is the same as that of ; that is, where . Clearly contains a subgroup isomorphic to , and since and is abelian, is normal in .

Suppose is not cyclic. By the induction hypothesis, there is a normal subgroup such that . By the Fourth (Lattice) Isomorphism Theorem, there is a normal subgroup such that . In particular, is a noncyclic group of order .

Now let be the -power map. By this previous exercise, is a homomorphism and in fact . Note that is not cyclic since is not cyclic. Thus, also by this exercise, has order or . Moreover, note that if and , then has order and thus has order . In particular, . So , and since is finite, in fact . Thus is characteristic in , hence normal in .

If , then is (isomorphic to) the elementary abelian group of order ; thus is normal in .

[With a hint from Alfonso Gracia-Saz’s notes] Suppose ; then in fact . Thus is the elementary abelian group of order , and . Let denote the set of all subgroups of order . Note that acts on by conjugation, and that . Now given , by the Orbit-Stabilizer Theorem, we have ; since is a -group, each orbit thus has order a power of . Since , there is thus one orbit of order and one orbit of order 1. Let be in the orbit of order 1. That is, is normal in , and by the Fourth Isomorphism Theorem, is normal. Finally, , and is elementary abelian. Thus has the desired properties.

Finally, suppose is a group which contains a unique subgroup of order . If is not cyclic, then by the above argument has a subgroup isomorphic to , yielding a contradiction because this subgroup contains many subgroups isomorphic to . Thus is cyclic.