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.

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.

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.

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.