# How many generators does a direct product of alternating groups need?

P. Hall gave a formula for the number of generators of $G^n$ for any finite simple group $G$. One famous example is the fact that $A_5^{19}$ is 2-generated, but $A_5^{20}$ is not. The question of computing $d(G^n)$ has extensively been studied, starting with the work of Wiegold. However, what we need is a bound for the direct product of different factors. More precisely, for $k_1<k_2<\dots<k_r$, how large can we take the exponents $e_i$, such that $A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r}$ is still 2-generated?

The closest related result I could find is due to Kravchenko and Petrenko (Some formulas for the smallest number of generators for finite direct sums of matrix algebras, Theorem 2.6), where the corresponding problem for full matrix algebra is dealt with. I would assume that the answer for simple groups is similar to the one for matrix algebras. The formula $d(A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r})=\max_i d(A_{k_i}^{e_i})$ might be too optimistic, but I guess that it is not far from the truth.

• I don't think $d(A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r})=\max_i d(A_{k_i}^{e_i})$ is optimistic. Any subgroup of $A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r}$ that projects onto each of the $r$ directs factors must be the whole group, and the result follows from that. – Derek Holt Mar 22 '17 at 12:19
• @DerekHolt: Thank you! Now I am feeling stupid. – Jan-Christoph Schlage-Puchta Mar 22 '17 at 13:01
• More generally, if the finite groups $G_1,\ldots,G_k$ have pairwise disjoint sets of composition factors, then $d(G_1 \times \cdots \times G_k) = \max_id(G_i)$. – Derek Holt Mar 22 '17 at 13:20
• see: mathoverflow.net/a/187774/1345 Given @DerekHolt's comment, I think this question is essentially a duplicate of this and other questions. Of course, there may be more that one can say about generating pairs for alternating groups, up to the automorphism group. – Ian Agol Mar 22 '17 at 18:45
• I was aware of the problem of powers, but did not think of using the composition series. – Jan-Christoph Schlage-Puchta Mar 22 '17 at 19:24

• as @DerekHolt points out, $d(A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r})=\max_i d(A_{k_i}^{e_i})$. This is a simple observation: if one has $d$ generators for $A_{k_i}^{e_i}$ for all $i$, this is a surjective homomorphism $\phi_i:F_d\to A_{k_i}^{e_i}$, where $F_d$ denotes the free group on $d$ generators. Then the diagonal $\phi_1\times \cdots\times \phi_r$ will have image generating $A_{k_1}^{e_1}\times\dots\times A_{k_r}^{e_r}$. To see this, note the subgroup it generates will surject each factor, and thus will have $A_{k_i}^{e_i}$ in its composition series. By uniqueness of the composition series (Jordan-Hölder Theorem), the group must therefore be a product of these factors.
• As you note, Philip Hall proved that the maximal number $n$ such that $G^n$ is $k$-generated is $h_k(G)$, the Eulerian function, which counts the number of $k$-generating sets for $G$, divided by $|Aut(G)|$. Thus, the problem is reduced to determining the number of $k$-generating sets for $G$ for all $k$ (assuming one knows $Aut(G)$). In the literature, this is usually expressed as the probability that $k$ elements generate, so is divided by $|G|^k$ to get a number $< 1$, denoted by $p_k(G)$.
• For $A_n$, we have $Aut(A_n)=S_n$ for $n\geq 7$. The best estimates on $p_2(A_n)$ appear to be
$$1−\frac{1}{n}−\frac{8.8}{n^2} ≤ p_2(A_n)<1−\frac{1}{n}−\frac{0.93}{n^2}.$$
Putting these facts together should give the maximal known possible value of $e_r$ that you desire.