The pth-Power Map (5.2.7)

Dummit and Foote Abstract Algebra, section 5.2, exercise 7:

Let $p$ be a prime and let $A = < x_1 > \times < x_2 > \times ... \times < x_n >$ be an abelian $p$-group, where $| x_i | = p^{\alpha_i} > 1$ for all $i$. Define the $p^{th}$-power map: $$φ : A → A~~~\text{by}~~~φ : x \mapsto x^p.$$ (a) Prove that $φ$ is a homomorphism.
(b) Describe the image and the kernel of $φ$ in terms of the given generators.
(c) Prove $\text{ker}~φ ≅ A/\text{im}~φ ≅ E_{p^n}$.

Proof: (a) $φ(ab) = (ab)^{p} = a^pb^p = φ(a)φ(b), ∀a,b∈A$.

(b) We can see that $< x_1^p > \times < x_2^p > \times ... \times < x_n^p > ≤ \text{im}~φ$, so we must prove the reverse inclusion. Let $a∈\text{im}~φ$, so rewrite $a$ thus: $$a = (a_1^p, a_2^p, ..., a_n^p) = ([x_1^{b_1}]^p, [x_2^{b_2}]^p, ..., [x_n^{b_n}]^p)$$ $$= ([x_1^p]^{b_1}, [x_2^p]^{b_2}, ..., [x_n^p]^{b_n})∈< x_1^p > \times < x_2^p > \times ... \times < x_n^p >$$ Therefore, $< x_1^p > \times < x_2^p > \times ... \times < x_n^p > = \text{im}~φ$.

Note that $< (x_1)^{p^{\alpha_1-1}} > \times < (x_2)^{p^{\alpha_2-1}} > \times ... \times < (x_n)^{p^{\alpha_n-1}} > ≤ \text{ker}~φ$. Prove the reverse inclusion: $$\text{if}~c∈\text{ker}~φ, \text{then}~c=(c_1, c_2, ..., c_n)=(x_1^{m_1}, x_2^{m_2}, ..., x_n^{m_n})$$ Where $(x_i^{m_i})^p=x_i^{m_ip}=1$ for all $i$, so that $p^{\alpha_i} \mid m_ip$, so $p^{\alpha_i-1} \mid m_i$, so $x_i^{m_i}=((x_i)^{p^{\alpha_i-1}})^{v_i}$ for some integer $v_i$. Thus, we have $< (x_1)^{p^{\alpha_1-1}} > \times < (x_2)^{p^{\alpha_2-1}} > \times ... \times < (x_n)^{p^{\alpha_n-1}}>~=\text{ker}~φ$.

(c) A small result to start: If $G$ is a finite abelian group such that for all $g∈G$, we have $g^p = 1$, then $G≅E_{p^n}$ for some $n$. Proof: Observe $G$'s elementary divisor decomposition. If any of its cyclic factors are not of order $p$, then there is an element of order neither $p$ nor $1$, a contradiction. Therefore $G≅Z_p^n=E_{p^n}$.

Now, for any $a∈A$, we have $a^p∈\text{im}~φ$, so for any $\bar{a}∈A/\text{im}~φ$, we have $\bar{a}^p=1$, so $A/\text{im}~φ$ is an elementary abelian group. Likewise, we have $k^p=1$ for all $k∈\text{ker}~φ$, so $\text{ker}~φ$ is an elementary abelian group. We have $A/\text{ker}~φ≅\text{im}~φ$, so that $|A/\text{im}~φ| = |\text{ker}~φ|$. All that remains is proving that $\text{ker}~φ$ is of rank $n$. We know that each of the group factors of $\text{ker}~φ$ has an order of $p$, so therefore $\text{ker}~φ$ is of order $p^n$ and thus isomorphic to $E_{p^n}$. $\square$