Let be a finite abelian group (written multiplicatively) and let be a prime. Let and . (I.e. and are the image and kernel of the -power map, respectively.
- Prove that . (Show that they are both elementary abelian and have the same order.)
- Prove that the number of subgroups of of order equals the number of subgroups of index . (Reduce to the case where is an elementary abelian -group.)
We begin with some lemmas.
Lemma 1: Let and be abelian groups and let be an isomorphism. Then there exists an isomorphism . Proof: Let denote the natural projection from to (for ). If , then for some . Then , so that . By the remarks on page 100 in D&F, there exists a unique group homomorphism such that . ( is injective) Suppose . Then , thus , so that , hence . Now since is surjective, for some . Thus , so that . Thus , so that is injective. ( is surjective) Let . Now for some , so that . Hence is surjective.
Lemma 2: Let and be abelian groups and let be an isomorphism. Then . Proof: If , then where . Now , so that , hence . Thus . Let . Then x^p = 1$. Since is surjective, for some . Now , and since is injective, . Thus , hence .
Lemma 3: Let be a finite abelian group of order , let be an integer, and let denote the -power homomorphism. If , then is an isomorphism. Proof: Since is finite, it suffices to show that is injective. Let . Then , so that divides . By Lagrange, divides . Thus , hence , and is injective.
Now to the main result. By Theorem 5 in the text, where is a -group of rank and does not divide . By a lemma to the previous exercise, , where each map is the -power map on , , and , respectively.
We have by the previous exercise.
Similarly, , again using the previous exercise.
Now the first conclusion follows using Lemmas 1 and 2.
We begin with some more lemmas.
Lemma 4: Let be the elementary abelian group of order , and suppose where . Finally let be a group. If is a mapping such that for each and for all , then extends uniquely to a group homomorphism . Proof: Every element of can be written uniquely as , where and . Define . is well defined since the -expansion of is unique, and is a homomorphism since the commute with one another. To see uniqueness, suppose is a group homomorphism such that for all . Then . Thus is unique.
Note that every subgroup of of order is contained in , so that in fact the order subgroups of and coincide. Now because is elementary abelian, every nonidentity element has order and thus generates an order subgroup. Each such subgroup is generated by elements; thus there are order subgroups in , thus in .
Let be a subgroup of index . Now let , say , and suppose . Then , so that . Now is a group of order , so that ; then , a contradiction. Thus . Moreover, by the Third Isomorphism Theorem, . By the Lattice Isomorphism Theorem, the index subgroups of correspond precisely to the index subgroups of . In particular, it suffices to count the order and index subgroups in an elementary abelian -group.
Now acts on by . We claim that this action is transitive. To that end, let be index subgroups. Now since and are elementary abelian, thus and where . Let be a bijection, and choose some and . Now extends by Lemma 4 to a homomorphism . Clearly is surjective, hence an isomorphism, so that . Moreover, we see that . Thus this action is transitive, and we have for all index subgroups .
Now let be an arbitrary index subgroup and let be an automorphism of which stabilizes . As above, for some set with . Moreover, if then since is maximal. By Lemma 4, is determined uniquely by is action on and . We see also that since stabilizes , we must have for each . Since and , there are distinct choices we can make for . Moreover, may be chosen arbitrarily so long as ; thus there are choices for . Thus in total contains elements.
By the Orbit-Stabilizer Theorem and Lagrange's Theorem, .
Thus the theorem is proved.