Proving a Certain Subgroup of $\mathbb{Q}/\mathbb{Z}$ is Divisible

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In an effort to compute the injective envelope of $\mathbb{Z}_p$ for $p$ prime, I need to show that the group $A \subset \mathbb{Q}/\mathbb{Z}$ generated by $\{1/p^r : r \in \mathbb{Z}^+\}$ is divisible. My proof follows; I'm just looking for a check.

Let $\sigma \in A$ be arbitrary and pick any nonzero $b \in \mathbb{Z}$. We wish to divide $\sigma$ by $b$. We can write $\sigma = [a/p^r]$ for some $a \in \mathbb{Z}$, $r \in \mathbb{Z}^+$. Suppose first that $p \nmid b$, so that $(b,p^r) = 1$. Then the linear congruence $bx \equiv a \pmod{p^r}$ has a solution, and so we have $b[x/p^r] = [bx/p^r] = [a/p^r] = \sigma$. Now suppose $p \mid b$, and write $b = cp^s$ where $p \nmid c$. By the argument above we can find $\sigma_1 \in A$ such that $c\sigma_1 = \sigma$. Writing $\sigma_1 = [a_1/p^r]$, we can let $\sigma_2 = [a_1/p^{r+s}]$ so that $p^s\sigma_2 = \sigma_1$. Then $b\sigma_2 = \sigma$ as desired.

Is this all good?