## Return to Answer

4 Added link to covers of An, retracted false conjecture

Note that $\text{Inn}(X)$ is isomorphic to $X/Z(X)$. So your requirement is equivalent to $X$ being simple modulo the centre. For example $SL_n(\mathbb F_{p^m})$ satisfies this, since the centre is given by the scalars and $PSL_n(\mathbb F_{p^m})$ is simple. A similar idea (see also Derek's answer) can give you explicit examples of such extensions with $A_n$ as the simple quotient.

More generally, central extensions of a given group $S$ (which you take to be simple) by the group $A$ are classified by $H^2(S,A)$. So if you fix $S$ and $A$, you can try computing this cohomology group and thereby deciding if there are any such extensions apart from the direct product. Note that while $H^2$ actually classifies extensions up to splitting, in this particular case it just classifies extensions up to your trivial example, since a semidirect product by the centre is a direct product.

A class of examples of the sort you are looking for is given by so-called quasi-simple groups. They are exactly the sorts of groups you want but with the additional requirement that $X'=X$. This is not so severe: indeed, already your requirement implies that $X'Z/Z \triangleleft X/Z$, which is simple. So either $X'\leq Z$ and then $X/Z\cong (X/X')\big/ (Z/X')$ is a quotient of $X/X'$, hence simple and abelian, which implies that $X$ is cyclic modulo the centre, hence abelian; or $X'Z = X$. So you are never too far away from $X'=X$. (I suspect that one can show that your groups are either quasi-simple or direct products. Proofs orcounterexamples Counterexamples will be have been gratefully received.)

3 Clarified the analysis of quasi-simple vs "simple modulo centre"

Note that $\text{Inn}(X)$ is isomorphic to $X/Z(X)$. So your requirement is equivalent to $X$ being simple modulo the centre. For example $SL_n(\mathbb F_{p^m})$ satisfies this, since the centre is given by the scalars and $PSL_n(\mathbb F_{p^m})$ is simple.

More generally, central extensions of a given group $S$ (which you take to be simple) by the group $A$ are classified by $H^2(S,A)$. So if you fix $S$ and $A$, you can try computing this cohomology group and thereby deciding if there are any such extensions apart from the direct product. Note that while $H^2$ actually classifies extensions up to splitting, in this particular case it just classifies extensions up to your trivial example, since a semidirect product by the centre is a direct product.

A class of examples of the sort you are looking for is given by so-called quasi-simple groups. They are exactly the sorts of groups you want but with the additional requirement that $X'=X$. This is not so severe: indeed, already your requirement implies that $X'Z/Z \triangleleft X/Z$, which is simple. So either $X'\leq Z$ and then $X/X'$ X/Z\cong (X/X')\big/ (Z/X')$is a quotient of$X/X'$, hence simple and abelian, hence which implies that$X$is cyclic modulo the centre, hence abelian, ; or$X'Z = X$. So you are never too far away from$X'=X$. (I suspect that one can show that your groups are either quasi-simple or direct products. Proofs or counterexamples will be gratefully received.) 2 Modified the reference to quasi-simple groups; added 27 characters in body Note that$\text{Inn}(X)$is isomorphic to$X/Z(X)$. So your requirement is equivalent to$X$being simple modulo the centre, also called "quasi-simple". For example$SL_n(\mathbb F_{p^m})$satisfies this, since the centre is given by the scalars and$PSL_n(\mathbb F_{p^m})$is simple. More generally, central extensions of a given group$S$(which you take to be simple) by the group$A$are classified by$H^2(S,A)$. So if you fix$S$and$A$, you can try computing this cohomology group and thereby deciding if there are any such extensions apart from the direct product. Googling Note that while$H^2$actually classifies extensions up to splitting, in this particular case it just classifies extensions up to your trivial example, since a semidirect product by the centre is a direct product. A class of examples of the sort you are looking for is given by so-called quasi-simplewill also give groups. They are exactly the sorts of groups you some more explicit exampleswant but with the additional requirement that$X'=X$. This is not so severe: indeed, already your requirement implies that$X'Z/Z \triangleleft X/Z$, which is simple. So either$X'\leq Z$and then$X/X'$is simple and abelian, hence$X$is cyclic modulo the centre, hence abelian, or$X'Z = X$. So you are never too far away from$X'=X\$. (I suspect that one can show that your groups are either quasi-simple or direct products. Proofs or counterexamples will be gratefully received.)

1