I have a abelian group $(G,+)$, and element $a \in G$ and some positive integer $n$.
I would like to add an element $b$ to $G$ such that $nb=a$. To be more precise, I want to construct an Abelian group $G'$ such that $G$ is a subgroup of $G'$ and $G'$ has an element $b$ such that $nb=a$.
What I am thinking Consider $$ G':= G \times \{0,1,2, \ldots, n-1 \} $$ with the operation $$ (g,k) \oplus (h, m)= \left\{ \begin{array}{cc} (g+h, k+m) & \mbox{ if } k+m <n \\ (g+h+a, k+m-n) & \mbox{ if } k+m \geq n \\ \end{array} \right. $$ This looks like adding the 2-digit "numbers $gk+hm$ with carrying.
My intuition tells me that $(G', \oplus)$ is indeed a group, and then $b=(0,1)$ has the desired property.
But it looks painfull to show that $G'$ is a group, and I would be surprised if this operation is not known ( maybe in the context of semi-direct products?).
Question Is this construction known? If yes, what is a good reference for it?
P.S. I think I've seen something like this in a group theory course I took more than 20 years ago, but not sure :)
You may try $$G'=\frac{G\oplus\mathbb{Z}}{\left<\left(a,-n\right)\right>}.$$ In fact, we can define an embedding $\phi:G\rightarrow G'$ by $$g\mapsto \left\{(g,0)+k(a,-n):k\in\mathbb{Z}\right\},\quad \text{i.e. }\;g\mapsto\overline{(g,0)},$$ and we will have $$\phi(a)=\overline{(a,0)}=n\cdot\overline{(0,1)}$$