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.

  • 2
    $\begingroup$ Just take the direct product of $G$ with any group of cardinality $\kappa$. $\endgroup$ – Derek Holt Dec 16 '17 at 12:21
  • 2
    $\begingroup$ 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

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 $G$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.