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I have done some research, and this paper http://www.heldermann-verlag.de/gcc/gcc02/gcc028.pdf seems to suggest that the Chevalley group $G(2,5)$ of order 5859000000 is a quotient of $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$. Can we check this, and if it is true, will it help with finding more quotients of the group?

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Yes, $G(2,5)$ is a quotient of your group $G$. -- We can find generators $a$ and $b$ of $G(2,5)$ satisfying the relations in a few seconds with GAP (so in particular there is no need to buy Magma for this!):

gap> G25 := SimpleGroup("G2(5)");
G2(5)
gap> cl2 := G25.1^G25;; # conjugacy class of involutions
gap> cl3 := G25.2^G25;; # one of the 2 conjugacy classes of elements of order 3
gap> repeat             # find the desired generators 
>      a := Random(cl2); b := Random(cl3);
>    until Order(a*b) = 7 and Order(Comm(a,b)) = 10 and G25 = Group(a,b);
gap> Order(a);
2
gap> Order(b);
3
gap> Order(a*b);
7
gap> Order(Comm(a,b));
10
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You could also do without charge with the Magma calculator! –  Derek Holt Nov 21 '13 at 11:58
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Yes, it's not hard to check! Just choosing random elements of $G_2(5)$ of order 2 and 3 in Magma, I very quickly found two matrices that satisfy the relations and generate $G_2(q)$:

[4 0 0 0 0 0 0]
[0 4 0 0 0 0 0]
[0 0 1 0 0 0 0]
[0 0 0 1 0 0 0]
[0 0 0 0 1 0 0]
[0 0 0 0 0 4 0]
[0 0 0 0 0 0 4]

[0 0 4 3 1 0 0]
[3 1 4 3 1 3 1]
[4 3 3 1 2 1 2]
[1 2 0 4 1 0 0]
[3 3 0 4 1 0 2]
[1 3 3 4 4 0 4]
[1 4 3 1 1 1 2]

A good reference for these types of calculations is W. Plesken and D. Robertz, Representations, commutative algebra, and Hurwitz groups, J. Algebra 300 (2006), 223-247. They find all characteristic zero representations of the Hurwitz group up to degree 7.

They say there that there are two 7-dimensional representations of $\langle a,b \mid a^2=b^3=(ab)^7=[a,b]^{10}=1 \rangle$ defined over quadratic extensions of the number field ${\mathbb Q}(\sqrt{5})$. That means that there are finite 7-dimensional representations over fields of order $p$ or $p^2$ or $p^4$ for almost all primes $p$. I don't know exactly what the images would be - probably linear or unitary groups for most primes - I can try and find out.

But in any case, there are definitely infinitely many finite simple images of the group.

Added: I have managed to compute a presentation of $G_2(5)$ by introducing a third generator $z$ and two long relations involving $y$ and $z$. It won't fit your idea of a simple presentation, but that's a limitation of computer calculations! The subgroup $\langle y,z \rangle$ has order 744000 and index 7875, so I got the presentation by first getting one for the subgroup and then doing coset enumeration over the subgroup.

< x,y,z |  x^2, y^3, (x*y)^7, (x,y)^10, z=x*y*x*y^-1*x,
  y^-1*z*y*z*y^-1*z*y*z*y*z*y*z*y^-1*z*y^-1*z*y*z*y^-1*z*y^-1*z*y^-1*z*y*z*
    y*z*y*z*y^-1*z*y^-1*z*y^-1*z*y*z*y^-1*z*y*z*y^-1*z* y*z*y^-1*z,
  y^-1*z*y*z*y^-1*z*y*z*y^-1*z*y*z*y^-1*z*y*z*y^-1*z*y*z*y^-1*z*y^-1*z*y^-1*
    z*y^-1*z*y*z*y*z*y^-1*z*y*z*y^-1*z*y*z*y^-1*z*y*z*y^-1*z*y^-1*z*y*z*y*z*
    y*z*y*z >;

Added: I have calculated explicitly the two 7-dimensional representations over number fields now, and tried reducing modulo some finite prime powers. The finite images all seem to be $G_2(p^e)$ for $e \le 4$. As well as $G_2(5)$,I've found $G_2(3^4)$, $G_2(7^2)$, $G_2(11^2)$, $G_2(13^4)$.

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Is there a simple presentation of the group as a quotient of G? –  Thomas Nov 21 '13 at 10:57
    
I don't know. The problem is that the group is too big to easily test computationally whether a candidate presentation is correct. –  Derek Holt Nov 21 '13 at 11:28
    
So, you're saying that G(2,49), G(2,81), G(2,121), and G(2,28561) are all quotients of G? –  Thomas Nov 22 '13 at 1:48
    
Yes, that's right. It looks as though $G(2,p^k)$ is a quotient of $G$ $k \le 4$ for all primes, possibly with one or two of exceptions (I don't think it works for $p=2$). –  Derek Holt Nov 22 '13 at 8:13
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