When is Inn(X) simple? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:27:43Z http://mathoverflow.net/feeds/question/46951 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46951/when-is-innx-simple When is Inn(X) simple? Makhalan Duff 2010-11-22T14:08:34Z 2010-11-24T08:23:58Z <p>Throughout, I restrict myself to finite groups.</p> <p>I'm interested in the question: can we characterize (or at least give some non-trivial examples or a family of non-trivial examples) the $X$'s such that $Inn(X)$ is simple? The obvious ones are $A \times G$ where $A$ is abelian and $G$ is simple, so I will consider those the "trivial" examples.</p> http://mathoverflow.net/questions/46951/when-is-innx-simple/46954#46954 Answer by Alex Bartel for When is Inn(X) simple? Alex Bartel 2010-11-22T14:29:00Z 2010-11-24T08:23:58Z <p>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 <a href="http://en.wikipedia.org/wiki/Covering_groups_of_the_alternating_and_symmetric_groups" rel="nofollow">extensions with $A_n$ as the simple quotient</a>.</p> <p>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.</p> <p>A class of examples of the sort you are looking for is given by so-called <a href="http://en.wikipedia.org/wiki/Quasisimple_group" rel="nofollow">quasi-simple</a> 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 <strong>and</strong> abelian, which implies that $X$ is cyclic modulo the centre, hence abelian; or $X'Z = X$. So you are never too far away from $X'=X$. (<s>I suspect that one can show that your groups are either quasi-simple or direct products. Proofs or</s> Counterexamples <s>will be</s> have been gratefully received.)</p> http://mathoverflow.net/questions/46951/when-is-innx-simple/47001#47001 Answer by Derek Holt for When is Inn(X) simple? Derek Holt 2010-11-22T20:57:52Z 2010-11-22T20:57:52Z <p>For every simple group $S$, there is a unique largest quasisimple group $\hat{S}$ such that $\hat{S}/Z(\hat{S}) \cong S$, and every quasisimple group with quotient $S$ is a quotient of $\hat{S}$ by a subgroup of $Z(\hat{S})$. Then $\hat{S}$ is known as the <em>covering group</em> of $S$, and $Z(\hat{S})$ is the <em>Schur Multiplier</em> (or <em>Multiplicator</em>) $M(S)$ of $S$. We also have $M(S) \cong H_2(S,\mathbb{Z})$.</p> <p>If $X$ is a group with $X/Z(X) \cong S$, then $X' \cap Z(X)$ is isomorphic to a quotient group of $M(S)$. It is possible that $X$ is neither quasisimple nor a direct product. For example, if we let $S = {\rm PSL}(2,q)$ for odd $q$, then $\hat{S} = {\rm SL}(2,q)$. Let $M(S) = \langle z \rangle$ with $|z|=2$ and $A = \langle y \rangle$ with $|y|=4$. Then the quotient group of $\hat{S} \times A$ by the subgroup $\langle zy^2 \rangle$ is neither quasisimple nor a direct product. (It is known as a <em>central product</em> of $M(S)$ and $A$.)</p>