The smallest integer $n$ such that there exists two non-isomorphic simple groups of order $n$, is $n=20160$ (namely for the groups $\mathrm{PSL}_3(\mathbb F _4)$ and $\mathrm{PSL}_4(\mathbb F _2)$). I read that there are infinitely many integer $n$ such that here exists two non-isomorphic simple groups of order $n$. I have two questions:

  1. Do you have a reference (if possible self contained, but that's probably too much to ask)?
  2. I suspect that it is "rare" to find such an integer. For instance if we denote by $a_k$ the orders of non-cyclic simple groups ($a_1=60$, $a_2=168$, $a_3=360$,....) and $b_k$ the integers such that there exists two non-isomorphic simple groups of order $b_k$, then I guess that $\displaystyle \lim\frac{b_k}{a_k}=+\infty$. Do you know if this is the case?


  • 7
    $\begingroup$ $\mathrm{P}\Omega(2\ell+1,q)$ and $\mathrm{PSp}(2\ell,q^2)$ have the same order and are nonisomorphic if $\ell\gt 2$. $\endgroup$ – Arturo Magidin Sep 19 '12 at 21:47
  • 1
    $\begingroup$ Note also that $20160$ is the order of the alternating group $A_{8}.$ The fact that the simple groups $A_{8}$ and ${\rm GL}(4,2)$ are isomorphic may be considered as rather exceptional, and I would call it non-obvious. $\endgroup$ – Geoff Robinson Sep 20 '12 at 0:24
  • 6
    $\begingroup$ Copy-paste from oeis.org/A109379: The first proof that there exist two nonisomorphic simple groups of order 20160 was given by the American mathematician Ida May Schottenfels (1869-1942): Ida May Schottenfels, Two Non-Isomorphic Simple Groups of the Same Order 20,160, Annals of Math., 2nd Ser., Vol. 1, No. 1/4 (1899), pp. 147-152. $${}$$ The orders for which there is more than one simple group are tabulated, with references, at oeis.org/A119648. $\endgroup$ – Gerry Myerson Sep 20 '12 at 0:44
  • $\begingroup$ @Arturo, Isn't it $\mathrm{PSp}(2\ell,q)$ instead of $\mathrm{PSp}(2\ell,q^2)$? $\endgroup$ – Portland Sep 20 '12 at 2:54
  • 4
    $\begingroup$ @Geoff: Here's a "geometric" proof. For $H = \{\sum x_i = 0\}$ in affine 8-space over $\mathbf{F}_2$, and $q = \sum_{i<j} x_i x_j$, $q|_H$ has defect line $L = \{x_1=\dots=x_8\}$. The quadratic space $(H/L,q)$ identifies $S_8$ with ${\rm{O}}_6(\mathbf{F}_2)$ through the $S_8$-action on affine 8-space preserving $H$, $L$, and $q$, so $A_8 = {\rm{SO}}_6(\mathbf{F}_2)$ as the unique index-2 subgroups. The isogeny ${\rm{SL}}_4 \simeq {\rm{Spin}}_6 \rightarrow {\rm{SO}}_6$ induces an isomorphism on $\mathbf{F}_2$-points, and ${\rm{SL}}_4(\mathbf{F}_2)={\rm{GL}}(4,2)$, so ${\rm{GL}}(4,2)=A_8$ $\endgroup$ – grp Sep 20 '12 at 4:39

Just to summarise the comments: the only nonisomorphic finite simple groups with the same orders are

  1. $A_8 \cong {\rm PSL}_4(2)$ and ${\rm PSL}_3(4)$ of order 20160.

  2. The groups ${\rm P \Omega}_{2n+1}(q)$ and ${\rm PSp}_{2n}(q)$ for all odd prime powers $q$ and $n \ge 3$. These have order

$$(q^{n^2} \Pi_{i=1}^n (q^{2i}-1))/2$$

For references, see Gerry Myerson's comment.

  • 2
    $\begingroup$ #2 is a "shadow" of the purely inseparable isogeny ${\rm{Spin}}_{2n+1} \rightarrow {\rm{Sp}}_{2n}$ in char. 2 that induces an isomorphism on $\mathbf{F}_{2^m}$-points for all $m>0$. Indeed, by Steinberg (or cohomological arguments over Spec($\mathbf{Z}$)), for a simply connected Chevalley group $G$ and a finite field $k$, $\#G(k)$ is a polynomial in $|k|$ depending only on the "type" of $G$ and not on char($k$), so equality for different types and all $q$ follows from equality as $q$ varies through powers of one prime (such as 2). In this sense, #1 seems more mysterious. $\endgroup$ – grp Sep 21 '12 at 10:08
  • 2
    $\begingroup$ Shameless Mathieu plug: Even more mysterious about #1: the smallest group containing subgroups isomorphic to both $A_{8}$ and $PSL_{3}(4)$ is the Mathieu group $M_{23}$. $PSL_{3}(4)$ is also beastly on its own because its outer automorphism group is large, its Schur multiplier is very large (order 48), and its Schur multiplier is related to the notoriously elusive Schur multiplier of $M_{22}$, in which $PSL_{3}(4)$ is a subgroup of index 22. To see $A_{8} \cong GL_{4}(2)$, one can consider how the stabilizer of an octad in $Aut S(5,8,24) = M_{24}$ acts on the 24 points. $\endgroup$ – DavidLHarden Sep 24 '12 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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