Problem 5.2.10 ($A/kA \simeq (\mathbb{Z}/ k \mathbb{Z})^n$ ($A$ abelian free group of rank $n$))

Question

Let $n$ and $k$ be positive integers and let $A$ be the free abelian group of rank $n$ (written additively). Prove that $A/kA$ is isomorphic to the direct product of $n$ copies of $\mathbb{Z}/ k \mathbb{Z}$ (here $kA = \{ka \mid a \in A\}$. [See Exercise 14, Section 1.]

Richard Ganaye · Accepted Answer

\begin{proof} By definition of a free abelian group of rank $n$ ,
$$A \simeq \mathbb{Z}^n.$$
Consider, as in Exercise 5.1.14, in additive notations now, the map
$$
\varphi
\left\{
\begin{array}{ccl}
\mathbb{Z}^n & 	o & (\mathbb{Z}/ k \mathbb{Z})^n\
(x_1,x_2,\ldots x_n) & \mapsto & ({[}x_1{]}_k, {[}x_2{]}_k,\ldots, {[}x_n{]}_k)
\end{array}
ight..
$$
Then 
\begin{align*}
(x_1,x_2,\ldots x_n) \in \ker(\varphi) & \iff {[}x_1{]}_k = {[}x_2{]}_k=\ldots =  {[}x_n{]}_k = [0]_k\
&\iff x_1 \equiv x_2 \equiv \cdots \equiv 0 \pmod k\
&\iff (x_1,x_2,\ldots x_n) \in (k\mathbb{Z})^n = k \mathbb{Z}^n.
\end{align*}
so
$$\ker(\varphi) =  k\mathbb{Z}^n.$$

Moreover $\varphi$ is surjective: if $(\alpha_1,\ldots,\alpha_n) \in (\mathbb{Z}/ k \mathbb{Z})^n$, there are integers $x_1,\ldots, x_n$ such that $\alpha_i = [x_i]_k$, and then $(\alpha_1,\ldots,\alpha_n)  = \varphi(x_1,x_2,\ldots x_n) $.

The First Isomorphism Theorem gives
 $$ \mathbb{Z}^n/ k\mathbb{Z}^n \simeq (\mathbb{Z}/ k \mathbb{Z})^n.$$

Since $A \simeq \mathbb{Z}^n$, this is equivalent to
$$A/kA \simeq  (\mathbb{Z}/ k \mathbb{Z})^n.$$

\end{proof}

Problem 5.2.10 ($A/kA \simeq (\mathbb{Z}/ k \mathbb{Z})^n$ ($A$ abelian free group of rank $n$))

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Problem 5.2.10 ($A/kA \simeq (\mathbb{Z}/ k \mathbb{Z})^n$ ($A$ abelian free group of rank $n$))

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