I'm sorry i don't get it.
Why i should add extra c and d letter if c = a^-1 = a and d = b^-1 = bb so i will get the same strings just by enumerate all strings of symbols consisting a and b ( i don't have problem with it)
next steep, as you wrote, should be remove redundant copies. I do it...
Here is my problem:
i have group defined by generators, like:
< a, b | a^2 = b^3 = (ab)^5 = 1 >.
eg. from http://for.mat.bham.ac.uk/atlas/v2.0/alt/A5/
i can't find algorithm to enumerate all element of group ( 60 in this example ) based on generators.
thanks for any help :)