Properties of the p-power map on abelian groups

Let p be a prime and let A = \langle x_1 \rangle \times \cdots \times \langle x_n \rangle be an abelian p-group, where |x_i| = p^{\alpha_i} > 1 for each i. Define the pth-power map \varphi : A \rightarrow A by \varphi(x) = x^p.

  1. Prove that \varphi is a homomorphism.
  2. Describe the image and kernel of \varphi in terms of the generators x_i.
  3. Prove that both \mathsf{ker}\ \varphi and A/\mathsf{im}\ \varphi have rank n (i.e. the same rank as A and prove that these groups are both isomorphic to the elementary abelian group E_{p^n}.

Without loss of generality, A has type (\alpha_1,\alpha_2,\ldots,\alpha_n).

  1. Let x,y \in A be arbitrary. By this previous exercise, \varphi(xy) = (xy)^p = x^py^p = \varphi(x)\varphi(y). Thus \varphi is a homomorphism.
  2. First we show that \mathsf{ker}\ \varphi = \prod_{i=1}^n \langle x_i^{p^{\alpha_i - 1}} \rangle.

    (\subseteq) Suppose \prod a_i \in \mathsf{ker}\ \varphi. Now \prod a_i = \prod x_i^{q_i} for some q_i, and \varphi(\prod a_i) = 1, so that x_i^{q_ip} = 1 for each i. Thus |x_i| = p^{\alpha_i} divides q_ip. We have q_i = mp^{\alpha_i - 1} for some m, so that a_i \in \langle x_i^{p^{\alpha_i - 1}} \rangle for each i; hence \prod a_i \in \prod_{i=1}^n \langle x_i^{p^{\alpha_i - 1}} \rangle.

    (\supseteq) If \prod a_i \in \prod_{i=1}^n \langle x_i^{p^{\alpha_i - 1}} \rangle, we have a_i = x_i^{mp^{\alpha_i - 1}} for some m, for each i. Then a_i^p = 1, so that (\prod a_i)^p = 1, hence \prod a_i \in \mathsf{ker}\ \varphi.

    Now we show that \mathsf{im}\ \varphi = \prod_{i=1}^n \langle x_i^p \rangle.

    (\subseteq) Suppose \prod a_i \in \mathsf{im}\ \varphi. Then there exists \prod b_i \in \prod_{i=1}^n \langle x_i \rangle with \varphi(\prod b_i) = \prod a_i. Now \prod b_i = \prod x_i^{m_i} for some m for each i, so that a_i = x_i^{m_ip}. Thus a_i \in \langle x_i^p \rangle, hence \prod a_i \in \prod_{i=1}^n \langle x_i^p \rangle.

    (\supseteq) Suppose \prod a_i \in \prod_{i=1}^n \langle x_i^p \rangle. Then for each i, there exists m_i with a_i = x_i^{pm_i}. Clearly then \varphi(\prod x_i^{m_i}) = \prod a_i, so that \prod a_i \in \mathsf{im}\ \varphi.

  3. We begin with some lemmas.

    Lemma 1: Let P = \langle x \rangle be a cyclic p-group of order p^\alpha > 1, and let \varphi : P \rightarrow P denote the p-power map. Then \mathsf{ker}\ \varphi \cong P/\mathsf{im}\ \varphi \cong Z_p. Proof: By the above argument, \mathsf{ker}\ \varphi = \langle x^{p^{\alpha-1}} \rangle and \mathsf{im}\ \varphi = \langle x^p \rangle. By this previous exercise, |\mathsf{ker}\ \varphi| = p and |\mathsf{im}\ \varphi| = p^{\alpha-1}, so that by Lagrange, |P/\mathsf{im}\ \varphi| = p. Now \mathsf{ker}\ \varphi and P/\mathsf{im}\ \varphi are both cyclic, hence \mathsf{ker}\ \varphi \cong P/\mathsf{im}\ \varphi \cong Z_p. \square

    Lemma 2: Let A_i be an abelian group, n an integer, let \varphi_i : A_i \rightarrow A_i denote the n-power map, let A = \prod_I A_i, and let \varphi : A \rightarrow A denote the n-power map. Then \varphi = \prod_I \varphi_i. Proof: If \prod a_i \in A, then \varphi(\prod a_i) = (\prod a_i)^n \prod a_i^n \prod \varphi_i(a_i) (\prod_I \varphi_i)(\prod a_i). \square

    Now to the main result.

    Note that \mathsf{ker}\ \varphi = \mathsf{ker}\ \prod_{i=1}^n \varphi_i = \prod_{i=1}^n \mathsf{ker}\ \varphi_i = \prod_{i=1}^n Z_p.

    Now \mathsf{im}\ \varphi = \mathsf{im}\ \prod_{i=1}^n \varphi_i = \prod_{i=1}^n \mathsf{im}\ \varphi_i. By this previous exercise, A/\mathsf{im}\ \varphi = (\prod_{i=1}^n \langle x_i \rangle)/(\prod_{i=1}^n \mathsf{im}\ \varphi_i) \cong \prod_{i=1}^n \langle x_i \rangle/\mathsf{im}\ \varphi_i \cong \prod_{i=1}^n Z_p, using Lemmas 1 and 2.

    Thus \mathsf{ker}\ \varphi \cong A/\mathsf{im}\ \varphi \cong \prod_{i=1}^n Z_p \cong E_{p^n}. In particular, \mathsf{ker}\ \varphi and A/\mathsf{im}\ \varphi have rank n.

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