I came across the following paragraph on pp.56 of Marshall Hall’s “The theory of groups” 1959 ed.

... the non-Abelian group of order 6 may be faithfully represented as a transitive permutation group on three letters and also on six letters ...

My question is,

In general, is there a theory about this? That is, given a non-Abelian group of order n, on what conditions it can be faithfully represented by different transitive permutation groups?

If the answer to the above question is yes, how do we call it? Multiple faithful representation? or something else?

Thanks in advance.


2 Answers 2


I don't know if there is a special name for this property, but I doubt it, as it is a very common property. Every finite group $G$ has a faithful transitive permutation representation on $|G|$ points, by Cayley's theorem. Given any subgroup $H$ of $G,$ there is a transitive permutation action of $G$ on the (say right) cosets of $H$ in $G$ – however, this action is not necessarily faithful. Its kernel is $\cap_{g \in G} g^{-1}Hg$, so the action is faithful just when $H$ contains no non-trivial normal subgroup of $G$. There are non-Abelian finite groups in which every proper non-trivial subgroup contains a proper non-trivial normal subgroup (such as (non-Abelian) Hamiltonian groups), but they are very much the exception, rather than the rule (to be precise, the exceptions are those non-Abelian groups in which every subgroup of prime order is normal). All other finite non-Abelian groups $G$ have a proper non-trivial subgroup $H$ such that $G$ acts faithfully and transitively on the $[G:H]$ right cosets of $H$ in $G.$

  • 1
    $\begingroup$ In fact, a finite group $G$ in which every subgroup of prime order is normal has to be solvable, but that requires some non-trivial group theory. Here's an outline: the derived subgroup $G^{\prime}$ has all subgroups of prime order central. This implies that for each odd prime $p$, $G^{\prime}$ has a normal $p$-complement. Thus $G^{\prime}$ has a normal Sylow $2$-subgroup, and when that is factored out, the quotient group is a nilpotent group of odd order. Thus $G^{\prime}$ is solvable, and so is $G.$ $\endgroup$ Feb 23, 2012 at 16:05
  • $\begingroup$ Do we have non-trivial examples other than M. Hall used? Thanks. $\endgroup$
    – Nobody
    Feb 24, 2012 at 2:12

Geoff Robinson's answer is perfect, of course, but I submit this answer to address your comment below Geoff's answer. (I can't write this as a comment because I want to include a picture.)

As Geoff points out, there are many non-trivial examples other than the example Hall presents. Consider the smallest non-abelian simple group, $A_5$. It is simple -- i.e. has no non-trivial normal subgroups -- so, for the reasons given by Geoff, every (non-trivial) permutation representation is faithful.

Here is a table (excerpted from my GapNotes.pdf) of all the permutation representations of $A_5$:

enter image description here

By the way, my profile picture happens to be the subgroup lattice of $A_5$. So, by looking at it, you can see the congruence lattice -- or "systems of imprimitivity," as the group theorists say -- of each of these representations; it's the interval above the given subgroup. The picture also makes it apparent that there are three primitive representations (since there are three conjugacy classes of maximal subgroups).

  • $\begingroup$ Geoff Robinson's answer is direct, thus I accept it. However, I really like this answer because I have something to work with. The only thing I can say is, thanks! $\endgroup$
    – Nobody
    Mar 8, 2012 at 9:46

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