$$< c^2,d^3,b^2c,ca^2,bda^3d^2,ab^3a^3b^3,b^3db^3d^2a^3 >$$
is a presentation of the group $SL(2,3)$.
According to GAP, the relators $c^2$ and $ab^3a^3b^3$ can be omitted, in other words,
$$< d^3,b^2c,ca^2,bda^3d^2,b^3db^3d^2a^3 >$$
is also a presentation of the group $SL(2,3)$.
I wanted to check this by hand :
What we know by the relators is :
- $d^3=e$
- $b^2c=e$
- $ca^2=e$
- $bda^3d^2=e$
- $b^3db^3d^2a^3=e$
This implies :
- $d^3=e$
- $a^2=b^2=c^{-1}$
- $bd=d(a^3)^{-1}$
- $b^3db^3=(a^3)^{-1}d$
Suppose, we could show $a^4=b^4=c^2=e$
Then, we would get $bd=da$ and $b^3db^3=ad$ implying $b^3d=adb$
Then we could conclude $ba=bd^2bd=bdabdd=bbdbdd=b^2abdd^2=b^2ab=a^3b$
because of $a=d^2bd$ and $db=a^3b^3d=cacbd=accbd=abd$ , the later equality follows from $ca=a^2a=a^3=aa^2=ac$. So finally we could derive $ab^3a^3b^3=ab^3(ba)^{-1}=ab^3(ab^3)^{-1}=e$.
But how can I derive $c^2=e$ or, equivalent , $a^4=b^4=e$ ?
I'm not sure whether there is a sufficiently nice solution to make a hand calculation feasible. Using a calculation in GAP (taking sufficiently many conjugates of relators to have subgroup generators for the kernel) I tried to express (e.g.) $c^2$ as a product of conjugates of relators, and got an expression that is a product of 800 factors. (If you really want this monster, send me a private email and I'll send it to you.) Presumably you could check this by hand.
While there is no guarantee that this is not missing a shorter expression, I wouldn't be surprised either if the expression you look for will have length of 50 or so (and thus still be infeasible to find by hand calculation).
(Even when simplifying (the generators a and c are redundant) I still get length 34.)