Problem 5.2.7 (Kernel and image of $x \mapsto x^p$ in an abelian $p$-group)

Proof. Here

where $Z_{p^{{\alpha }_i}}=⟨x_i⟩$.

(a)

Since $A$ is abelian, for all $a,b\in A$, $$\phi (\mathit{𝑎𝑏})={(\mathit{𝑎𝑏})}^p=a^p{b}^p=\phi (a)\phi (b),$$

so $\phi$ is a homomorphism.

(b)

If $a=(a_1,a_2,\dots ,a_n)\in A$, then $a_i=x_i^{k_i}$ for some integer $k_i$ $(1\le i\le n)$. Then $$\begin{array}{llll}\hfill \phi (a)& =a^p& \hfill & \\ \hfill & =(x_1^{k_{1}p},x_2^{k_{2}p},\dots ,x_p^{k_{n}p})& \hfill & \\ \hfill & \in ⟨x_1^p⟩\times ⟨x_2^p⟩\times \cdots \times ⟨x_n^p⟩.& \hfill & \end{array}$$

Therefore

$$\mathrm{im}(\phi )\le ⟨x_1^p⟩\times ⟨x_2^p⟩\times \cdots \times ⟨x_n^p⟩.$$

Conversely, if $y=(y_1,y_2,\dots ,y_n)\in ⟨x_1^p⟩\times ⟨x_2^p⟩\times \cdots \times ⟨x_n^p⟩$, then $y_i={(x_i^p)}^{k_i}$ for some integer $k_i$ ( $1\le i\le n$), thus

$$y={(x_1^{k_i},x_2^{k_i},\dots ,x_n^{k_i})}^p\in \mathrm{im}(\phi ),$$

so

$$\begin{array}{lll}\hfill \mathrm{im}(\phi )=⟨x_1^p⟩\times ⟨x_2^p⟩\times \cdots \times ⟨x_n^p⟩.& & \hfill \text{(1)}\end{array}$$

If $a=(a_1,a_2,\dots ,a_n)=(x_1^{k_1},x_2^{k_2},\dots ,x_n^{k_n})\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $x_i^{k_{i}p}=1$ for all $i$. Since $|x_i|=p^{{\alpha }_i}$, we obtain $p^{{\alpha }_i}\mid k_{i}p$, thus

$$p^{{\alpha }_i-1}\mid k_i.$$

We may write $k_i=q_i{p}^{{\alpha }_i-1}$ for some integer $k_i$, thus

$$a=(x_1^{q_1{p}^{{\alpha }_1-1}},x_2^{q_2{p}^{{\alpha }_2-1}},\dots ,x_n^{q_n{p}^{{\alpha }_n-1}})\in ⟨x_1^{p^{{\alpha }_1-1}}⟩\times ⟨x_2^{p^{{\alpha }_2-1}}⟩\times \cdots \times ⟨x_n^{p^{{\alpha }_n-1}}⟩.$$

This shows

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\le ⟨x_1^{p^{{\alpha }_1-1}}⟩\times ⟨x_2^{p^{{\alpha }_2-1}}⟩\times \cdots \times ⟨x_n^{p^{{\alpha }_n-1}}⟩.$$

Conversely, if

$$a=(a_1,a_2,\dots ,a_n)=(x_1^{k_1},x_2^{k_2},\dots ,x_n^{k_n})\in ⟨x_1^{p^{{\alpha }_1-1}}⟩\times ⟨x_2^{p^{{\alpha }_2-1}}⟩\times \cdots \times ⟨x_n^{p^{{\alpha }_n-1}}⟩,$$

then $p^{{\alpha }_i-1}\mid k_i$ for all $i$, so $p^{{\alpha }_i}\mid k{p}_i$. Therefore ${(x_i^{k_i})}^p=1$, thus $a^p=1$ and $a\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$. This proves

$$\begin{array}{lll}\hfill \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=⟨x_1^{p^{{\alpha }_1-1}}⟩\times ⟨x_2^{p^{{\alpha }_2-1}}⟩\times \cdots \times ⟨x_n^{p^{{\alpha }_n-1}}⟩.& & \hfill \text{(2)}\end{array}$$

(c)

Possibly permuting the factors $⟨x_i⟩$ of $A$, we may assume ${\alpha }_1\ge {\alpha }_2\ge \cdots \ge {\alpha }_n$, so that $$A=Z_{p^{{\alpha }_1}}\times Z_{p^{{\alpha }_2}}\times \cdots \times Z_{p^{{\alpha }_n}}\text{where }{p}^{{\alpha }_{i+1}}\mid p^{{\alpha }_i}(1\le i<n)$$

is the decomposition of $A$ in invariant factors, and so $n$ is the rank of $A$.

By (1) and Exercise 5.1.14,

$$\begin{array}{llll}\hfill A∕\mathrm{im}(\phi )& =(⟨x_1⟩\times ⟨x_2⟩\times \cdots \times ⟨x_n⟩)∕⟨x_1^p⟩\times ⟨x_2^p⟩\times \cdots \times ⟨x_n^p⟩& \hfill & \\ \hfill & \simeq (⟨x_1⟩∕⟨x_1^p⟩)\times (⟨x_2⟩∕⟨x_2^p⟩)\times \cdots \times (⟨x_n⟩∕⟨x_n^p⟩).& \hfill & \\ \hfill & & \hfill \end{array}$$

Moreover, for every $i$, $⟨x_i⟩∕⟨x_i^p⟩$ is generated by $\overline{x_i}=x_i⟨x_i^p⟩$, where ${\overline{x_i}}^p=⟨x_i^p⟩=\bar{1}$, and $\overline{x_i}\ne 1$, since $x_i\notin ⟨x_i^p⟩$ (otherwise $|x_i|<p^{{\alpha }_i}$). Hence $⟨x_1⟩∕⟨x_1^p⟩\simeq Z_p$, so

$$\begin{array}{lll}\hfill A∕\mathrm{im}(\phi )\simeq Z_p\times Z_p\times \cdots \times Z_p=Z_p^n.& & \hfill \text{(3)}\end{array}$$

So the rank of $A∕\mathrm{im}(\phi )$ is $n$.

Similarly, by (2), since $|x_i^{p^{{\alpha }_i-1}}|=p$,

$$\begin{array}{lll}\hfill \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\simeq Z_p\times Z_p\times \cdots \times Z_p=Z_p^n.& & \hfill \text{(4)}\end{array}$$

Therefore the rank of $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ is $n$.

So $\ker <!--\; FUNCTION\; APPLICATION\; -->\phi$ and $A∕\mathrm{im}\phi$ have rank $n$ (i.e., have the same rank as $A$).

By (3) and (4),

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\simeq A∕\mathrm{im}(\phi )\simeq Z_p^n,$$

are isomorphic to the elementary group, $E_{p^n}$, of order $p^n$.