A problem on finite abelian group

Question

I recently started studying group theory and there's a problem I encountered in the book Undergraduate Algebra - Serge Lang that I'm currently unable to solve:
Let $(A,+)$ be a finite abelian group of order $n$, and let $n=p_1^{r_1}...p_s^{r_s}$ be its prime power factorization, the $p_i$ being distinct.
a) Show that $A=A_1\oplus ... \oplus A_s$ where every element $a \in A_i$ satisfies $p_i^{r_i}a=0$
b) Prove that $\#(A_i)=p_i^{r_i}$
I've solved part a) of the problem but I'm stuck at b).
My idea is to try and prove that either $\#(A_i) \leq p_i^{r_i}$ or $\#(A_i) \geq p_i^{r_i}$ for all $1 \leq i \leq s$, then use the fact that $\#(A)=\#(A_1)...\#(A_s)$ to conclude but I don't know how to.

Answer 1

Lemma: If $A$ is a $p^n$-torsion finite Abelian group for some prime $p$ (and natural $n$), then the order of $A$ is a power of $p$.

Proof: In fact, this holds even without the hypothesis that $A$ is Abelian (cf. Cauchy’s theorem, but the Abelian case is very easy by strong induction (the base case that $A$ is trivial) and quotienting by cyclic subgroups. $\Box$

---

Okay, now simply remark that by Lagrange’s theorem (and the right-exactness of tensor), $$A\ \simeq\ A/nA\ \simeq\ A\otimes\mathbb{Z}/n$$ and also that by the Chinese remainder theorem, $$\mathbb{Z}/n\ \simeq\ \bigoplus_{i}\mathbb{Z}/p_{i}^{r_{i}}$$ so that $$A\ \simeq\ A\otimes\mathbb{Z}/n\ \simeq\ A\otimes \left(\bigoplus_{i}\mathbb{Z}/p_{i}^{r_{i}}\right)\ \simeq\ \bigoplus_{i}\left(A\otimes\mathbb{Z}/p_{i}^{r_{i}}\right)\ \simeq\ \bigoplus_{i}A/p_{i}^{r_{i}}A$$ whence $$\prod_{i} p_{i}^{r_{i}}\ =\ n\ =\ \left|A\right|\ =\ \prod_{i} \left|A/p_{i}^{r_{i}}A\right|$$ and as each $A/p_{i}^{r_{i}}A$ is $p_{i}^{r_{i}}$-torsion, it it follows from Lemma that it has order $p_{i}^{r_{i}}$, completing the proof. $\blacksquare$