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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 :)

Elías Guisado Villalgordo
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    Let $G = H \times \mathbb Z_k$. –  Aug 23 '21 at 08:31
  • @Bungo, what if we truly require $H\subseteq G$ as sets? –  Aug 23 '21 at 08:44
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    @CAB I'm not even sure what that means, I'm afraid. – David A. Craven Aug 23 '21 at 08:49
  • @DavidA.Craven, for $k=2$ it means this construction: https://math.stackexchange.com/q/3614000/943729 –  Aug 23 '21 at 08:52
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    @CAB I'm afraid I think your question doesn't really make much sense. Groups are not defined canonically. Any set of size 3 can have a group structure placed on it, for example. So to say that $H$ is a subset of $G$ just requires you to move the elements from your original set into the direct product. – David A. Craven Aug 23 '21 at 08:56
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    But if we are given $H$ defined with a specific underlying set, and we know that the group $G$ contains an isomorphic copy of $H$, then we can just extend the underlying set of $H$ to one of size $|G|$ and transfer the group structure from $G$ to that set (and I am not making any finiteness assumptions here). But the answer to the question is obviously yes – Derek Holt Aug 23 '21 at 09:09

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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 $\operatorname{Sym}(H\sqcup Q)$.

user1729
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