Minimal generating set in case of finite non-abelian groups Finding a minimal generating set of a finite group is difficult but in the case of abelian groups using the fundamental theorem of finite abelian groups, we can find it easily. Moreover, corresponding to each $k\in \mathbb N$, we have an abelian group such that the cardinality of a minimal generating set is $k$ (take rank $k$ abelian groups). In the case of non-abelian groups, we know that symmetric group on $n$-letters $S_{n}$ can be generated by two elements. Can you give an example of a finite non-abelian group that cannot be generated by two elements? In general, can we find a nonabelian group corresponding to each $k>1$ such that the cardinality of a minimal generating set is $k$?
Thank you.
 A: You are using the following definition:

A minimal generating set for a group $G$ is a set $S$ such that $\langle S\rangle = G$, but for every proper subset $T$ of $S$, $\langle T\rangle\neq G$.

As Derek Holt points in comments, this makes an affirmative answer to your question easy: the set
$$\{(1,2), (1,3),\ldots,(1,k-1), (1,k), (1,k+1)\}$$
is a minimal generating set with $k$ elements for $S_{k+1}$.
In fact, $S_n$ has minimal generating sets of sizes $k$, $2\leq k\leq n-1$: the set $\{(1,2),(1,3),(1,4),\ldots,(1,k),(k,k+1,\ldots,n)\}$ is easily verified to a minimal generating set.
Moreover, as noted in comments here, it is a theorem that if a finite group $G$ has a minimal generating set with $k$ elements, and a minimal generating set with $\ell\gt k$ elements, then it has minimal generating sets with $m$ elements for all $m$, $k\leq m\leq\ell$. This is Tarski's Irredundant Basis Theorem. See references cited in this MathOverflow question about the values of $D(G)$, the size of the largest possible minimal generating set for $G$.
On the other hand, given a finite group $G$, if we let $d(G)$ denote the cardinality of the smallest minimal generating set for $G$ (also called the "rank of $G$" or the "generating rank of $G$"), you might ask whether for any $k>1$ there exists a finite nonabelian group such that $d(G)=k$. Again, as noted in the comments, an extra-special $p$-group ($p\gt 2$) of order $p^{2n+1}$ has generating rank $2n$, and then taking the direct product with a cyclic group of order $p$ gives a group with generating rank $2n+1$, so the answer is "yes" here as well.
