Suppose $G$ is a $k$-generated finite group. Is there always a set of $k$ elements which generate the group and have a unimodal counting function?

**Background:**
The counting function, $f(n)$, is a function whose value at $n$ is the number of elements of $G$ of length $n$.

The length of an element $x$, relative to a given generating set, is the length of the shortest word, made up only of elements of the generating set, which is equal to $x$. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion.

A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence $\{s_1, s_2,\ldots,s_n\}$ is unimodal if there exists a $t$ such that $s_1\leq s_2\leq\cdots\leq s_t$, and $s_t \geq s_{t+1} \geq \cdots\geq s_n$. (Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html)

**Discussion:**
As pointed out in one of the comments, taking $k$ to be the number of elements in the group will trivially give unimodal growth. The question is more interesting if one considers values of k which are less than the number of elements in the group. In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact, nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function.

Here is an example of a group with non-unimodal growth. The two generators are the permutations:

$a = \{7, 9, 2, 3, 6, 5, 1, 8, 4\}$

$b = \{6, 8, 7, 5, 2, 1, 3, 4, 9\}$

In this notation $a$ is the permutation which takes 1 into 7, 2 into 9, 3 into 2,...

The counting function for this pair of generators is:

1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2.