# Is every group a nontrivial subgroup of a larger group?

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.

• Just take the direct product of $G$ with any group of cardinality $\kappa$. – Derek Holt Dec 16 '17 at 12:21
• 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. – Arnaud D. Dec 16 '17 at 12:28
• @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 \$? – Gustavo Dec 16 '17 at 12:32
• @ArnaudD. How do you define $\ X \sqcup Y \$? – Gustavo Dec 16 '17 at 13:03
• 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. – MJD Dec 16 '17 at 13:42

## 1 Answer

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 $G$.

• 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. – hmakholm left over Monica Dec 16 '17 at 13:43
• Right. Thanks. That's a good point. – Asaf Karagila Dec 16 '17 at 14:18
• This is great answer. Thanks. – Gustavo Dec 16 '17 at 14:53