Extending an infinite simple group

Question

Maybe the question does not fit here.

Yesterday in my logic course, I presented a nice example about an application of model theory to group theory. The example is due to Hodges and as following: For any infinite simple group $G$, there is a countable simple group $H\subseteq G$.

Then a logician asked whether the upward version is true. i.e. $\mathbf{ \mbox{is it true that for every infinite simple group $H$ and cardinal }\kappa\geq |H|, \mbox{there is a simple group }G\supseteq H \mbox{ with } |G|=\kappa?}$

To give a positive answer to the question, it is sufficient to prove that every infinite simple group $H$ is a proper subgroup of a simple group $G$.

I was told by an algebraist that the question for finite simple groups has a negative answer.

$\mathbf{Remark}$: The question was already answered by Derek Holt. He proved that one may find such a $G$ with $|G|=2^{\kappa}$ for some cardinal $\kappa$. Then by Skolemization including $H$, one may obtain a full answer.

Answer 1

I will make my comment into an answer. It is a standard result (see for example Chapter 8 of Dixon & Mortimer's book on Permutation Groups) that, for an infinite cardinal $\kappa$, the only normal subgroups of the symmetric group ${\rm Sym}(\Omega)$ on a set $\Omega$ of cardinality $\kappa$ are the subgroups ${\rm Sym}(\Omega,\lambda)$, consisting of those permutations of $\Omega$ with support strictly less than $\lambda$, for all cardinals $\lambda$ with $\aleph_0 \le \lambda \le \kappa$, and also the subgroup ${\rm Alt}(\Omega)$ of even permutations with finite support.

So, if $H$ is any infinite group with cardinality $\kappa$ then the standard regular embedding of $G$ into ${\rm Sym}(H)$, in which the image of each non-identity element has support equal to $H$, induces an embedding into the simple group ${\rm Sym(H)}/{\rm Sym}(H,\kappa)$, which has cardinality $2^\kappa$.