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 OrbitStabilizer Theorem and Lagrange's Theorem, .
Thus the theorem is proved.
Comments
in the last parag.,How does the last equality follow? where does p^(t1) factor go?
Good question. I’ll try to find the problem later today.
Oh sorry i see my misteke there is no problem,everything is ok.
Maybe you could explain it to me.🙂 I am still convinced there is a mistake.
i mean the last calculations. you have already modified them now its true i guess.
by the way in some solutions to the problems as in this one you say “By the remarks on page … ” what does “page …” refer to?how can i get to these pages?
Sorry the page references are to Dummit and Foote’s “Abstract Algebra”, 3rd edition.
I did fix a mistake in the expressions which count the automorphisms, but unless I am missing something I think it is still wrong. The strategy seems to be okay, but that pesky factor doesn’t go away. Either I’m making an arithmetic error, or something is counted wrong.
in the paragraph just before the last one, the product over i should start from i=0 to i=t2 as stated in 4 lines above that line.
nevertheless in the last paragraph you start the product from i=0 but this time the general term over that product should be “p^(t1)p^(t2i)” not “p^(t1)p^(t1i)”.this solves the problem i guess now.
by the way i wonder where did you find such an important theorem?is this from a special book or an article or?
thanks for all your goodness!
Thanks! All of these problems come from Dummit and Foote’s “Abstract Algebra”, 3rd edition. Eventually I will start pulling problems from other books too.
During the course of trying to solve this problem, I found out that in fact, in an elementary abelian group of order , the number of subgroups of order and is the same for all . Moreover, in a certain sense, this result is a generalization of the binomial theorem. The Wiki page on qanalogs gives more information on this.
Hey, I just thought I’d let you know. There is a short proof for the number of indexed subsets of the elementary abelian group of order assuming you know the rudiments of linear algebra. Namely, if is an elementary abelian group of order then it’s easy to see that is (canonically) a vector space over and then note that subgroups and subspaces coincide (since multiplication corresponds to addition). In particular, the subgroups of of order are precisely the subspaces of of dimension and those of index are those of order . But, it is a simple fact of linear algebra that if is an dimensional space then the number of dimensional subspaces is equal to –it’s a simple counting argument about choosing linearly independent subsets (it’s on pg. 412 of Dummit and Foote).
And of course, plugging in and gives the desired result (not only that their equal, but that they are equal to .
Thanks!