# Finite group with at least 3 generators?

The answer to this question could be trivial, but usually I do not work in group theory.

Is there a finite nonabelian group $G$ that is generated by $S$ where $S$ is a minimal generating set of $G$, and $3\leq |S|\leq |G|-1$? (Other than the direct product of nonabelian or abelian groups)

What I'm looking here is finite nonabelian group that satisfies the above inquality. Of course the obvious answer is the direct product of certain finite nonabelian or abelian group, but I'm not interested in it.

For the infinite nonabelian groups we can take the free group of n-generators (with n≥3).

I notice most of the finite nonabelian groups like dihedral, symmetric group,.. does not satisfy the above condition.

Any help will be useful!

• The nonabelian group $G=S_3$ is generated by the set $S=\{(1\ 2),(1\ 3),(2\ 3)\}.$ Did you leave out some condition you wanted? – bof Mar 7 '16 at 9:53
• @bof It seems clear that a minimal generating set is sought for. – Tobias Kildetoft Mar 7 '16 at 9:53
• @TobiasKildetoft That was my guess too, but I really like people to say what they mean. – bof Mar 7 '16 at 9:55
• Any non-abelian simple group can be generated by two elements, so if you want an example that contains no direct product the closest I can think of is a semidirect product... – DonAntonio Mar 7 '16 at 10:26
• There is no such thing as the generating set for a group. You were explicitly asked if you were looking for a minimal generating set, and you said no. You also said that $S_3$ was a good example, even though that is in fact a dihedral group. – Tobias Kildetoft Mar 7 '16 at 10:37

There is an example of order $16$ which is not a direct product, namely the central product of $D_8$ (dihedral of order $8$) and $C_4$.
What about $\;Q_8\times C_2\;,\;\;Q_8=$ the quaternion group?