I'm trying to calculate a table of all simple hurwitz groups of order less than 10^7. None of the tables I found went further than 10^6, so I decided to use the tables of all simple groups up to 10^7 (which is easy to find), and then remove the one which are not hurwitz. Using some of the papers I have read on hurwitz groups I have managed to reduce the list down to only the projective special linear groups of degree 2 (that are hurwitz), the Janko groups J1 and J2 (which I know are hurwitz) and four others, which I am not sure about. The steinberg groups 2A(3, 9), 2A(2, 49), 2A(2, 64), and the chevalley group C(3, 2). Which of these four groups are hurwitz (I suspect none of them)?

Added later: I now know all hurwitz groups with orders less than 10^10. These are the PSL(2,q) groups (which satisfy the conditions for it to be hurwitz), J1, J2, 3D(4, 2), He, G(2, 5), and 2G(2, 27). There is only one group that is unknown up to 10^12, and that is C(4, 2) (which is isomorphic to B(4, 2)).

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    $\begingroup$ A Hurwitz group is a finite group that occurs (up to isomorphism) as the automorphism group of a Riemann surface of genus $g$, and has the maximum possible order $84(g-1)$ for such a group. (from groupprops.subwiki.org/wiki/Hurwitz_group) $\endgroup$
    – YCor
    Commented Feb 4, 2015 at 8:21
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    $\begingroup$ Your notation for the $^2A_2$'s is a bit puzzling - I'm not sure what you mean here. However in general $^2A_2(q)$ are 3-dimensional unitaries and so are not Hurwitz. In any case, for a summary of the state of play (in 2000) you should look at the paper by Martino, Tamburini and Zalesski here: mathematik.uni-bielefeld.de/LAG/man/021.ps.gz $\endgroup$
    – Nick Gill
    Commented Feb 4, 2015 at 8:37
  • $\begingroup$ Also $C(3,2)$ presumably means $C_3(2) = {\rm PSp}(6,2)$, which is not Hurwirz. $\endgroup$
    – Derek Holt
    Commented Feb 4, 2015 at 9:02
  • $\begingroup$ heldermann-verlag.de/gcc/gcc02/gcc028.pdf is a more recent reference. $\endgroup$
    – Derek Holt
    Commented Feb 4, 2015 at 9:13
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    $\begingroup$ $B(2,7) \cong {\rm P}\Omega(5,7) \cong {\rm PSp}(4,7)$, which is not Hurwitz. All Hurwitz groups with representations of degree up to $7$ have been determined. $\endgroup$
    – Derek Holt
    Commented Feb 4, 2015 at 11:46

1 Answer 1


In the paper

M.C. Tamburini and M. Vsemirnov, Irreducible $(2,3,7)$-subgroups of ${\rm PGL}_n(F)$, $n \le 7$, J. Algebra 300 (2006), 339–362

the Hurwitz groups with absolutely irreducible projective representations of degrees up to $7$ over any field are determined (although the results in the smaller dimensions (up to $5$ I think) were not new.

Here is a list of the simple groups that arise - I hope I have copied this correctly!

Added later: I am sorry, I misunderstod the paper. This is not a complete list - this is a list of so-called rigid triples - and the authors say that they will complete the classification in a leter paper, which has now appeared in J. Algebra 321 (2009), no. 8, 2119–2138. In the second paper, they have not completely identified the groups that arise in dimension $7$, but they remark that some of the groups $G_2(q)$ arise - these had been found earlier by Malle.

$n=2$: ${\rm PSL}(2,p^m)$, where $m=1$ if $p \equiv 0,\pm 1 \mod 7$, and $m=3$ otherwise.

$n=3$ and $n=4$: nothing new.

$n=5$: ${\rm PSL}(5,p^m)$ and ${\rm PSU}(5,p^m)$ with $p \ne 5$ for certain values of $m$. See the survey paper by Conder for details.

$n=6$: ${\rm PSL}(6,p^m)$, with $p \ne 3$ and $m$ odd, and ${\rm PSU}(6,p^m)$, with $p \ne 3$ and $m$ even where, in both cases, $m$ is the order of $p$ mod $9$.

$n=7$: ${\rm PSL}(7,p^m)$, with $p \ne 7$ and $m$ odd, and ${\rm PSU}(7,p^m)$, with $p \ne 7$ and $m$ even where, in both cases, $m$ is the order of $p$ mod $49$.

Added later: I checked with a computer calculation that none of the three groups of order less than $10^9$ that you are uncertain about are Hurwitz. These are $C_2(7) = {\rm PSp}(4,7)$, $D_4(2) = {\rm P \Omega}^+(8,2)$ and $^2D_4(4) = {\rm P \Omega}^-(8,2)$. The first of these has a projective representation of degree $4$ and is covered by known results: none of the $4$-dimensional symplectic groups are Hurwitz. As far as I know, the $8$-dimensional orthogonal groups are not covered by published results, but it is very likely that somebody has dome these calculations already! The computer checks I used are more or less brute force, and they work easily for groups of order up to $10^9$, but will start to become impractical with group orders much higher than that.

I checked also that ${\rm He}$ is not an image of $(2,3,7;10)$.

  • $\begingroup$ A query: according to the ATLAS, $J_1$ is a subgroup of $G_2(11)$ which in turn is a subgroup of $O_7(11)$ (I think? Not totally sure about that.) But Thomas says above that $J_1$ is a Hurwitz group. So shouldn't $J_1$ have cropped up in your list? Or is its omission covered by the caveat about dimension $7$? $\endgroup$
    – Nick Gill
    Commented Feb 4, 2015 at 15:43
  • $\begingroup$ On another front: is it true that all groups of order at most $10^7$ (or $10^8$ or $10^9$) have projective representations of dimension at most $7$? If there are some that don't, these would not be covered by your answer, right? $\endgroup$
    – Nick Gill
    Commented Feb 4, 2015 at 15:44
  • $\begingroup$ That can't be right: ${\rm PSL}(2,q)$ has order about $q^3$ and its smallest projective representation has dimension about $q/2$, so we already get beyond dimension $7$ with a simple group of order $<10^4$. Did you mean to impose some additional condition? $\endgroup$ Commented Feb 4, 2015 at 20:55
  • $\begingroup$ According to the paper by Conder, the groups 3D(4, q) are all hurwitz except when q is 4 or a power of 3. Therefore the group 3D(4, 2) is hurwitz. That leaves three simple groups with orders less than 10^9 that are uncertain. $\endgroup$
    – Thomas
    Commented Feb 5, 2015 at 2:51
  • $\begingroup$ @NickGill I would guess that $J_1$ is one of the groups that arise in dimension $7$ in the second (2009) paper of Tamburini and Vsemirnov. The authors make no attempt to identify them, but merely remark that they must include the groups $G_2(q)$ found by Malle. In the case of $q=11$, the group in question is a proper subgroup of $G_2(11)$, and I would think that $G_2(11)$ itself is not Hurwitz. $\endgroup$
    – Derek Holt
    Commented Feb 5, 2015 at 9:21

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