MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

I have found it necessary to define the following property:

Consider a finite set $X$ and a group $H$ of permutations of $X$. Suppose for every normal subgroup $K$ of $H$ that $K$ acts faithfully on every $K$-orbit.

The property I have is that $G$ acts on $X$, not necessarily faithfully, inducing a permutation group $H$ as above.

Is there a standard name for the property possessed by $H$?

Certainly primitive permutation groups are examples for $H$, but so are fixed-point-free permutation groups for instance. On the other hand, $S_n \wr S_k$ acting on $nk$ points in the usual imprimitive way is not an example.

I think 'quasi-primitive' is already taken, and means 'every normal subgroup is transitive'. This is strictly stronger than the property I am talking about.

share|cite|improve this question
If $H$ is transitive, it seems to be the case that $K$ is either faithful on each of its orbits, or is faithful on none of them,so in tht case, you seem to have just excluded the case that $K$ does not act faithfull on any orbit (which does happen, as your wreath product example shows). – Geoff Robinson Jul 5 '11 at 18:02
Yes, that is more or less the point. Another way of saying it is that given $1 < L \unlhd K \unlhd H$ then $L$ is not contained in any point stabiliser of $H$. – Colin Reid Jul 5 '11 at 19:18
up vote 2 down vote accepted

So this question is not left unanswered, I should add that I decided to call them 'subprimitive'. The context is a construction of some examples of hereditarily just infinite profinite groups, which has now appeared in my paper 'Inverse system characterizations of the (hereditarily) just infinite property in profinite groups' in Bulletin of the LMS:

share|cite|improve this answer

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.