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Let $G$ be a group and $\kappa$ be a cardinal number strictly bigger than $|G|$.

(1) Is there a group $H$ for which $G$ is a nontrivial subgroup of $H$?

(2) Is there a group $H$ for which $G$ is a nontrivial subgroup of $H$ and $ \ |H|= \kappa \ $?


Let $ \ \mu: H \times H \to H \ $ be the binary operation of the group $H$. That is, $(H, \mu)$ is the group. I say that a group $(J, \nu)$ is a subgroup of $(H, \mu)$ if, and only if,

(i) $ \quad J \subset H \ $;

(ii) $ \quad \mu|_{J \times J} = \nu \ $.

Here I can only use ZFC set theory. That is, equality is "set theory equality" and identifications or isomorphisms are strictly prohibited.


This question appeared because in a semigroup $S$ without identity it is very easy to add an extra element (set) $ \ e \notin S \ $ such that $ \ S \cup \{ e \} \ $ is a semigroup with identity $e$, just defining by hand what the operation do on it. I wonder if this is possible with other elements and other structures as well.

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    $\begingroup$ Just take the direct product of $G$ with any group of cardinality $\kappa$. $\endgroup$ – Derek Holt Dec 16 '17 at 12:21
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    $\begingroup$ I'm not an expert in ZFC, but it seems to me that given an injective function $f:A\to B$, you can always define $B'=(B\setminus f(A))\sqcup A$ and $f':A\to B'$ as the inclusion. Then you have a bijection $\theta:B\to B'$ such that $\theta \circ f=f'$. So any solution "up to isomorphism" can be turned into a "strict" solution. $\endgroup$ – Arnaud D. Dec 16 '17 at 12:28
  • $\begingroup$ @DerekHolt Do you mean that the underlying set of the direct product is the cartesian product $ \ G \times H \ $? How do you construct this cartesian product in a way that $ \ G \subset G \times H \ $? $\endgroup$ – Gustavo Dec 16 '17 at 12:32
  • $\begingroup$ @ArnaudD. How do you define $ \ X \sqcup Y \ $? $\endgroup$ – Gustavo Dec 16 '17 at 13:03
  • $\begingroup$ Coincidentally, just yesterday I was reading about the construction that takes a ring (which may not have an identity) and extends it to a ring with identity. The construction is similar to the one that turns a semigroup into a monoid. $\endgroup$ – MJD Dec 16 '17 at 13:42
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Once you know that the answer work "up to isomorphism", then the answer is positive in the set theoretic sense as well.

Suppose that $(H,*_H)$ is isomorphic to a subgroup of $(G',*_{G'})$. Namely, there is a monomorphism $\varphi\colon H\to G'$. Let $G$ be the set $H\cup(\{H\}\times(G'\setminus\operatorname{range}(\varphi)))$. We can replace $\{H\}$ by $\{x\}$ for which $H\cap\{x\}\times G'=\varnothing$, and of course there is a proper class of such $x$'s.

Now define $\Phi\colon G\to G'$ as follows: $$\Phi(x)=\begin{cases}\varphi(x) & x\in H\\ y & x=(H,y)\end{cases},$$ namely either $\varphi(x)$ or the projection onto $G'$. And define $$x*_Gy=\Phi^{-1}(\Phi(x)*_{G'}\Phi(y)).$$

It is not hard to check that $(G,*)$ is a group and that $H$ is indeed a subgroup of $G$. Of course, there are no restrictions on the cardinality of $G$.

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  • $\begingroup$ Note that this requires that $G'$ is disjoint from $H$, but that is of course easy to arrange. If it isn't, then first replace $G'$ by $G'\times\{H\}$, for example. $\endgroup$ – hmakholm left over Monica Dec 16 '17 at 13:43
  • $\begingroup$ Right. Thanks. That's a good point. $\endgroup$ – Asaf Karagila Dec 16 '17 at 14:18
  • $\begingroup$ This is great answer. Thanks. $\endgroup$ – Gustavo Dec 16 '17 at 14:53

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