That is the question: if we are given a group $H$ and a non-negative integer $k$, is it always possible to construct a new group $G$ such that $H$ is (isomorphic to) a subgroup of $G$ with index $[G:H]=k$?
This question arises from this other answer: https://math.stackexchange.com/a/364064. On that thread, OP asks whether every connected groupoid is an action groupoid. On the linked response, Omar's affirmative argument needs a positive answer to the question I have here posed.
In the linked post, the required $k$ is taken to be an arbitrary cardinal $k=\aleph$.
I don't know that much group theory so I don't know how to address the problem. If we were to try $G$ to be a direct product/direct sum of copies of $H$ and consider $H$ to be one of the factors/addends, that only would solve some cases, and same with the free product.
Any help or suggestions will be appreciated :)
Let $H$ be any group and $Q$ be a group of cardinality $k$. Set $G:=H\times Q$. Then $H$ has index $|Q|=k$ in $G$, i.e. $[G:H]=k$, as required. This works for all cardinals $k$ (using the fact that there are groups $Q$ of every cardinal; see here).
For example, if $k\in\mathbb{N}$ then take $Q=\mathbb{Z}/k\mathbb{Z}$.
The comments are talking about doing this is in some sort of "more concrete" way. We can use permutation groups to do this: we may assume that $H$ and $Q$ have disjoint underlying sets, then via Cayley's theorem we have that $H$ embeds into the permutation group $\operatorname{Sym}(H)$ and $Q$ embeds into the permutation group $\operatorname{Sym}(Q)$, and then $H\times Q$ embeds into the permutation group $\operatorname{Sym}(H\sqcup Q)$.