Properties of the image and kernel of the p-power map on a finite abelian group

Let A be a finite abelian group (written multiplicatively) and let p be a prime. Let A^p = \{ a^p \ |\ a \in A \} and A_p = \{ x \in A \ |\ x^p = 1 \}. (I.e. A^p and A_p are the image and kernel of the p-power map, respectively.

  1. Prove that A/A^p \cong A_p. (Show that they are both elementary abelian and have the same order.)
  2. Prove that the number of subgroups of A of order p equals the number of subgroups of index p. (Reduce to the case where A is an elementary abelian p-group.)

  1. We begin with some lemmas.

    Lemma 1: Let A and B be abelian groups and let \theta : A \rightarrow B be an isomorphism. Then there exists an isomorphism \theta^\prime : A/A^p \rightarrow B/B^p. Proof: Let \pi denote the natural projection from Z to Z/Z^p (for Z \in \{A,B\}). If x \in A^p, then x = y^p for some y \in A. Then (\pi \circ \theta)(x) = (\pi \circ \theta)(y^p) = \pi(\theta(y)^p) = B^p, so that A^p \leq \mathsf{ker}\ (\pi \circ \theta). By the remarks on page 100 in D&F, there exists a unique group homomorphism \theta^\prime : A/A^p \rightarrow B/B^p such that \theta^\prime \circ \pi = \pi \circ \theta. (\theta^\prime is injective) Suppose xA^p \in \mathsf{ker}\ \theta^\prime. Then \theta^\prime(\pi(x)) = 1, thus (\theta^\prime \circ \pi)(x) = 1, so that (\pi \circ \theta)(x) = 1, hence \theta(x) \in B^p. Now since \theta is surjective, \theta(x) = \theta(y)^p for some y \in A. Thus x = y^p, so that x \in A^p. Thus \mathsf{ker}\ \theta^\prime = 1, so that \theta^\prime is injective. (\theta^\prime is surjective) Let xB^p \in B/B^p. Now xB^p = (\pi \circ \theta)(y) for some y \in A, so that xB^p = \theta^\prime(\pi(y)). Hence \theta^\prime is surjective. \square

    Lemma 2: Let A and B be abelian groups and let \theta : A \rightarrow B be an isomorphism. Then \theta[A_p] = B_p. Proof: (\subseteq) If x \in \theta[A_p], then x = \theta(y) where y \in A_p. Now y^p = 1, so that \theta(y)^p = 1, hence x^p = 1. Thus x \in B_p. (\supseteq) Let x \in B_p. Then x^p = 1$. Since \theta is surjective, x = \theta(y) for some y \in A. Now \theta(y^p) = \theta(y)^p = x^p = 1, and since \theta is injective, y^p = 1. Thus y \in A_p, hence x \in \theta[A_p]. \square

    Lemma 3: Let G be a finite abelian group of order n, let m be an integer, and let \varphi : G \rightarrow G denote the m-power homomorphism. If \mathsf{gcd}(m,n) = 1, then \varphi is an isomorphism. Proof: Since G is finite, it suffices to show that \varphi is injective. Let x \in \mathsf{ker}\ \varphi. Then x^m = 1, so that |x| divides m. By Lagrange, |x| divides n. Thus |x| = 1, hence x = 1, and \varphi is injective. \square

    Now to the main result. By Theorem 5 in the text, A \cong A_1 \times A_2 where A_1 is a p-group of rank t and p does not divide |A_2|. By a lemma to the previous exercise, \varphi = \varphi_1 \times \varphi_2, where each map is the p-power map on A_1 \times A_2, A_1, and A_2, respectively.

    We have \mathsf{ker}\ \varphi = \mathsf{ker}(\varphi_1 \times \varphi_2) = (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2) \cong Z_{p^t} \times 1 \cong Z_{p^t} by the previous exercise.

    Similarly, (A_1 \times A_2)/\mathsf{im}\ \varphi = (A_1 \times A_2)/(\mathsf{im}\ \varphi_1 \times \mathsf{im}\ \varphi_2) \cong (A_1/\mathsf{im}\ \varphi_1) \times (A_2/\mathsf{im}\ \varphi_2) \cong Z_{p^t} \times 1 \cong Z_{p^t}, again using the previous exercise.

    Now the first conclusion follows using Lemmas 1 and 2.

  2. We begin with some more lemmas.

    Lemma 4: Let E_{p^k} be the elementary abelian group of order p^k, and suppose E_{p^k} = \langle S \rangle where |S| = k. Finally let G be a group. If \overline{\varphi} : S \rightarrow G is a mapping such that |\overline{\varphi}(s)| = p for each s \in S and \overline{\varphi}(s) \overline{\varphi}(t) = \overline{\varphi}(t) \overline{\varphi}(s) for all s,t \in S, then \overline{\varphi} extends uniquely to a group homomorphism \varphi : E_{p^k} \rightarrow G. Proof: Every element of E_{p^k} can be written uniquely as x = s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k}, where s_1 \in S and 0 \leq a_k < p. Define \varphi(x) = \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}. \varphi is well defined since the S-expansion of x is unique, and is a homomorphism since the \overline{\varphi}(s) commute with one another. To see uniqueness, suppose \psi : E_{p^k} \rightarrow G is a group homomorphism such that \psi(s) = \varphi(s) for all s \in S. Then \psi(x) = \psi(s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k}) = \psi(s_1)^{a_1} \psi(s_2)^{a_2} \cdots \psi(s_k)^{a_k} = \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k} = \varphi(x). Thus \varphi is unique. \square

    Note that every subgroup of A of order p is contained in \mathsf{ker}\ \varphi, so that in fact the order p subgroups of A and \mathsf{ker}\ \varphi coincide. Now because \mathsf{ker}\ \varphi \cong Z_{p^t} is elementary abelian, every nonidentity element has order p and thus generates an order p subgroup. Each such subgroup is generated by p-1 elements; thus there are (p^t-1)/(p-1) order p subgroups in \mathsf{ker}\ \varphi, thus in A.

    Let B \leq A be a subgroup of index p. Now let x \in A^p, say x = y^p, and suppose x \notin B. Then y \notin B, so that A = \langle y \rangle B. Now A/B is a group of order p, so that (yB)^p = xB = B; then x \in B, a contradiction. Thus A^p \leq B. Moreover, by the Third Isomorphism Theorem, [A/A^p : B/A^p] = [A:B] = p. By the Lattice Isomorphism Theorem, the index p subgroups of A/A^p \cong Z_p^t correspond precisely to the index p subgroups of A. In particular, it suffices to count the order and index p subgroups in an elementary abelian p-group.

    Now \mathsf{Aut}(Z_p^t) acts on \mathcal{M} = \{ M \leq Z_p^t \ |\ [Z_p^t : M] = p \} by \varphi \cdot M = \varphi[M]. We claim that this action is transitive. To that end, let M_1, M_2 \leq Z_p^t be index p subgroups. Now M_1 \cong M_2 \cong Z_p^{t-1} since M_1 and M_2 are elementary abelian, thus M_1 = \langle S_1 \rangle and M_2 = \langle S_2 \rangle where |S_1| = |S_2| = t-1. Let \overline{\psi} : S_1 \rightarrow S_2 be a bijection, and choose some t_1 \in Z_p^t \setminus M_1 and t_2 \in Z_p^t \setminus M_2. Now \overline{\psi} \cup \{(t_1,t_2)\} extends by Lemma 4 to a homomorphism \psi : Z_p^t \rightarrow Z_p^t. Clearly \psi is surjective, hence an isomorphism, so that \psi \in \mathsf{Aut}(Z_p^t). Moreover, we see that \psi[M_1] = M_2. Thus this action is transitive, and we have \mathsf{Aut}(Z_p^t) \cdot M = \mathcal{M} for all index p subgroups M.

    Now let M \leq Z_p^t be an arbitrary index p subgroup and let \varphi be an automorphism of Z_p^t which stabilizes M. As above, M = \langle S \rangle for some set S with |S| = t-1. Moreover, if x \in Z_p^t \setminus M then Z_p^t = \langle S \cup \{x\} \rangle since M is maximal. By Lemma 4, \varphi is determined uniquely by is action on S and x. We see also that since \varphi stabilizes M, we must have \varphi(s) \in M for each s \in S. Since M \cong Z_p^{t-1} and \varphi|_M \in \mathsf{Aut}(M), there are \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i}) distinct choices we can make for \varphi[S]. Moreover, \varphi(x) may be chosen arbitrarily so long as \varphi(x) \notin M; thus there are p^t - p^{t-1} = p^{t-1}(p-1) choices for \varphi(x). Thus in total \mathsf{Stab}(M) contains (p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i}) elements.

    By the Orbit-Stabilizer Theorem and Lagrange's Theorem, |\mathcal{M}| = (\prod_{i=0}^{t-1} (p^t - p^{t-1-i}))/((p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})) = (\prod_{i=1}^t (p^t - p^{t-i}))/((p-1) \prod_{i=1}^{t-1}(p^{t} - p^{t-i})) = (p^t-1)/(p-1).

    Thus the theorem is proved.

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  • quadro  On March 8, 2011 at 2:37 am

    in the last parag.,How does the last equality follow? where does p^(t-1) factor go?

  • nbloomf  On March 8, 2011 at 7:59 am

    Good question. I’ll try to find the problem later today.

  • quadro  On March 9, 2011 at 4:07 am

    Oh sorry i see my misteke there is no problem,everything is ok.

    • nbloomf  On March 9, 2011 at 10:42 am

      Maybe you could explain it to me. 🙂 I am still convinced there is a mistake.

  • quadro  On March 14, 2011 at 6:31 am

    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?

    • nbloomf  On March 14, 2011 at 9:19 am

      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 p^{t-1} factor doesn’t go away. Either I’m making an arithmetic error, or something is counted wrong.

  • quadro  On March 14, 2011 at 4:34 pm

    in the paragraph just before the last one, the product over i should start from i=0 to i=t-2 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^(t-1)-p^(t-2-i)” not “p^(t-1)-p^(t-1-i)”.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!

    • nbloomf  On March 15, 2011 at 8:32 am

      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 p^t, the number of subgroups of order p^k and p^{t-k} is the same for all k. Moreover, in a certain sense, this result is a generalization of the binomial theorem. The Wiki page on q-analogs gives more information on this.

  • Alex Youcis  On August 25, 2011 at 12:44 am

    Hey, I just thought I’d let you know. There is a short proof for the number of p-indexed subsets of the elementary abelian group of order p^n assuming you know the rudiments of linear algebra. Namely, if A is an elementary abelian group of order p^n then it’s easy to see that A is (canonically) a vector space over \mathbb{F}_p and then note that subgroups and subspaces coincide (since multiplication corresponds to addition). In particular, the subgroups of A of order p are precisely the subspaces of A of dimension 1 and those of index p are those of order n-1. But, it is a simple fact of linear algebra that if V is an n-dimensional \mathbb{F}_q-space then the number of k-dimensional subspaces is equal to \displaystyle \frac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{(q^{k}-1)(q^{k}-q)\cdots (q^k-q^{k-1})}–it’s a simple counting argument about choosing k linearly independent subsets (it’s on pg. 412 of Dummit and Foote).

  • Alex Youcis  On August 25, 2011 at 12:47 am

    And of course, plugging in q=p and k=1,n-1 gives the desired result (not only that their equal, but that they are equal to (p^n-1)(p-1)^{-1}.

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