## 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 $p$th-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 $n$.