# Is there any algorithm to find the minimal generating set for $A_{n}$

I know that finite simple groups can be generated by two elements.(See this question on MO)

So as a specific example, Take Alternating group $$A_{n}$$, $$n>4$$.

We also know that $$A_{n}$$ is $$(2,3)$$ generated.

I want to write a program, which takes a positive integer($$n>4$$) as input and return two elements of $$A_{n}$$ one having order $$2$$ and other having order $$3$$, also these two elements generate the whole group.

But I am not able to write an algorithm.

Any hints.

• $A_n$ is not $(2,3)$-generated for $n=6,7$ and $8$. If I wanted to do this I would just choose random elements of order $2$ and $3$ having minimal numbers of fixed points. You will find generators very quickly that way. – Derek Holt Nov 13 '16 at 11:27

As for algorithm, just take even involution with maximal support and try to find an even element of order 3 such that union of that partitions is just $(n)$; this pair will generate the whole group. That "greedy" algorithm will suffice if $n > 9$. Take that maximal involution $I = (12)(34)\dots(4k-1, 4k)$; take element $T$ of the form $(235)(679)..(4l+2, 4l+3, 4l+5), 4l+5 < 4k$ "connecting" all but maybe one 2-cycle. Now you have remaining fixed points (from 0 to 3) and 0 or 1 leftover 2-cycle. Number of holes in support of $T$ intersected with support of $I$ is at least $k+1$. Now arrange leftovers in pairs and connect it with support of $I$ adding 0 to 2 3-cycles to $T$ taking up no more than 3 holes. That "algorithm" (it admits closed form, obviously) is even 4-periodic in $n$. I'm not paying attention to primitivity, so this can be true only by coincidence.
Upd: there's whole article about explicit $(2,3)$ generators for $A_n$ and $S_n$