Groups generated by a binary operation on a set that contains one member of each conjugacy I was recently trying to solve the following problem:

Let $G$ be a finite group. Let $S\subset G$ be a set containing exactly one element of each conjugacy class of $G$. Prove that $S$ generates $G$.

I had the idea to have a look at the binary (conjugacy) operation defined as follows:
$$S\bullet S = \{s_1s_2s_1^{-1}\mid s_1,s_2\in S\}$$
and then construct the powers of this operation as
$$ S^n = \{s_1s_2\dots s_{n-1} s_n s_{n-1}^{-1}\dots s_1^{-1}\mid s_1,s_2,\dots,s_n\in S\}.$$
I don't think this operation is the right tool to solve the original problem, but it led me to an interesting question:
Suppose for some $n$ it is $S\bullet S^{n} = S^{n}$. What restrictions do we have to impose on the group $G$ and/or on $S$ in order for $S^{n}=G$?
For example, if for all $s\in S^{n}$ the inverse $s^{-1}$ is in $S^{n}$, this means that for some element $g\in S^{n}$ the following equation has to hold
$$gsg^{-1}=s^{-1}\implies gs = (gs)^{-1}$$
This means, that there exists an element of order 2 and thus the order of the group is even.
For some small-sized groups I have examined it worked, e.g. $S_3$ and the quaternion group.
The next step would then be to find the smallest $n$ for which $S^{n} = G$.
 A: The answer to your question is that, if $S^n=S^{n+1}$, then $S^n=G$, with no additional assumption.
I'm going to assume that we have already answered the original question. (Somebody posted a proof earlier, but then deleted it because that didn't answer your question.) In particular, $\langle S\rangle =G$. Note that, since the group is finite, in fact every element of $G$ is a product of elements of $S$ (with no need to use inverses).
Since $G$ is finite, the sequence $(S,S^2,\ldots)$ must eventually stabilise. Let us call $T$ the set to which it stabilises. Now, elements of $T$ are obtained by taking an element $s$ of $S$, and conjugating by an element $t$ which is a product of elements of $S$. But we saw above that every element of $G$ is a product of elements of $S$, so $t$ can be any element of $G$, and $S$ covers all the conjugacy classes, so $T=G$.
Finally, it is obvious that if $S^n=S^{n+1}$, then $S^n=T$.
Note that this approach uses the answer to the original question, so it's not useful to answer that question, but it does answer your question.
