$G$ is a permutation group. So, $G < S_n$ where $S_n$ is a symmetric group acts on $n$ object.

$G$ is not isomorphic to any symmetric or alternating group, i.e. $G \neq S_t , A_t$ for $1 <t \leq n$ where $S_t , A_t$ are symmetric group and alternating group respectively.


$k$ is the size of minimal generating set of group $G$.

How many minimal generating set of group $G$ are possible with size $k$?

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    $\begingroup$ I think this question is hopelessly broad. We know from Cayley's theorem that $G$ could be any group (except symmetric and alternating groups, by your exclusion). I suspect nobody can characterize the number of minimal generating sets of a generic group. You might want to play around with GAP (possibly through Sage) for some data, but I'm willing to bet it'll be all over the map. $\endgroup$ – pjs36 Jun 5 '16 at 1:15

The assumption $G \le S_n$ seems irrelevant. Every group embeds into a symmetric group, so $G$ might just as well be an arbitrary finite group.

If $|G|=m$ then the answer is clearly at most $\binom{m}{k}$ and in many (but not all) groups it is more than half of this number.

For example, for all finite nonabelian simple groups $G$, we have $k=2$ and the number of minimal generating sets is at least $53\binom{m}{k}/90$ with equality when $G=A_6$ (although I note that for some unknown reason you are not counting alterning or symmetric groups).

But for $G=A_5^{19}$, we still have $k=2$ but the number is much smaller as a proportion of $|G|$, and if you chose pairs of elements, you would be unlikely to find a pair of generators in the lifetime of the universe.

Cyclic groups are also interesting, where $k=1$ and the number of generators is $\Phi(m)$. Note that $\Phi(m)/m$ can be arbitrarily small.


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