Quotients of abelian groups - residual finiteness and elements of order $p$

Question

Suppose $A$ is an abelian group and $\pi$ is a set of prime numbers. A $\pi$-number is a product of primes from $\pi$.

Assume that for each $p \in \pi$, $A_p = \{a \in A : \exists i\in\mathbb{N} \text{ s.t } a^{p^i} = 0\}$ has finite exponent.

Assume also that $A$ is $\pi$-reduced; there are no non-trivial subgroups of $A$ which are $\pi$-divisible. That is, for any $H \leq A$ there is $h \in H$ and $m$ a $\pi$-number such that for any $x \in H$, $x^m \neq h$.

Let $j \in \mathbb{N}$, $p \in \pi$ and $m = p^jn$ a $\pi$-number where $n$ is relatively prime to $p$.

why is $A/A^m$ residually finite?
why does $A^{p^j}/A^m$ have no element of order $p$?

Here is the context from Infinite Soluble Groups:

@halrankard thanks, edited. I edited the second question and took it off for a few minutes. — Mariah, Aug 18 '20 at 13:48
Ok I see, it makes sense now. I'll delete the previous comments — halrankard, Aug 18 '20 at 13:49

score 1 · Accepted Answer · answered Aug 18 '20 at 22:05

$A/A^m$ is an abelian group of finite exponent (specifically, exponent dividing $m$), and every abelian group of finite exponent is a direct sum of cyclic groups and in particular residually finite. See for instance the answers at infinite abelian group where all elements have order 1, 2, or 4.

Since every element $a\in A/A^m$ satisfies $a^m=1$, every $p^j$th power $a^{p^j}$ satisfies $(a^{p^j})^n=1$. Thus the order of $a^{p^j}$ divides $n$, and so cannot be $p$. That is, $A^{p^j}/A^m$ has no elements of order $p$.

Thanks for that reference! The second question, I don't know why it confused me.. — Mariah, Aug 18 '20 at 22:17

Quotients of abelian groups - residual finiteness and elements of order $p$

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