MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Throughout, I restrict myself to finite groups.

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.

share|cite|improve this question
I'm not confident that any reasonable characterization of this class of finite groups exists, but groups of Lie type give lots of nontrivial examples starting with special linear groups over most finite fields. – Jim Humphreys Nov 22 '10 at 14:18
P.S. My comment may be an overstatement, but at any rate the groups in question are as noted by others just the quotients of simple groups (somewhat known) by their centers. – Jim Humphreys Nov 22 '10 at 23:11

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 or Counterexamples will be have been gratefully received.)

share|cite|improve this answer
Ah! I didn't know about quasi-simple groups. You say it's equivalent to quasi-simple? What about the requirement there that $X'=X$? In any case, it gives a nice family of examples. – Makhalan Duff Nov 22 '10 at 14:35
I first thought that this requirement was automatically satisfied, if your group isn't a direct product, but now I am not so sure. I will modify the post. But "simple modulo the centre" is definitely the term you are looking for. – Alex B. Nov 22 '10 at 14:58
mmm... Why are you saying that $X/X'$ is simple in the case that $X' \leq Z$? Wouldn't $Z/X'$ usually be a non-trivial normal subgroup of $X/X'$? – Makhalan Duff Nov 22 '10 at 15:42
I have tried to make it clearer. – Alex B. Nov 22 '10 at 16:10
Thanks! It's much clearer now. – Makhalan Duff Nov 22 '10 at 16:11

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 covering group of $S$, and $Z(\hat{S})$ is the Schur Multiplier (or Multiplicator) $M(S)$ of $S$. We also have $M(S) \cong H_2(S,\mathbb{Z})$.

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 central product of $M(S)$ and $A$.)

share|cite|improve this answer
If Z(X) has no elements of prime-squared order, must X' be a direct factor? It seemed like H^2(X/Z(X),p) would control it, and that it would not allow anything exciting. Maybe this also just follows by thinking about the definition of central product long enough? – Jack Schmidt Nov 22 '10 at 22:05
In that case (assuming $X$ finite of course) $Z(X)$ is a direct product of elementary abelian groups, so don't we get $Z(X) = (Z(X) \cap X') \times A$ for some $A$ and then $X = X' \times A$? – Derek Holt Nov 22 '10 at 22:18
Thanks! That makes the situation very clear. In general Z(X)∩X′ need not be a summand of Z(X), but of course it is in the elementary abelian case. – Jack Schmidt Nov 22 '10 at 23:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.