Q:

Is The Degree of the Minimal Monomial of a Group Equal to Its Exponent? 

Suppose $G$ is a group with identity $e$ such that $x^n=e$ for all $x\in G$ and such that no smaller $n$ satisfies this.  If, for some constants $a_i\in G$ it holds for all $x$ that
$$a_1\cdot x\cdot a_2\cdot x\cdot a_3\cdot x\cdot \ldots\cdot a_k \cdot x = e$$
then does $n$ necessarily divide $k$?  Put another way, if we consider the above to be a "monomial" term, defining its degree to be the number of times it multiplies by $x$, is it necessary that the exponent of the group divides the degree of any monomial sending all elements to the identity? 
This is clearly true of abelian groups, because then the above expression collapses to
$$x^k=\prod_{i=1}^ka_i^{-1}$$
for all $x$, implying $x^k=y^k$ for all $x,y\in G$, and setting $y=e$ yields that the above is equivalent to $x^k=e$ for all $x$, which does imply that the exponent of the group divides $k$.
Is the statement true in the non-abelian case too? 

A:

Counter-example: The exponent of the symmetric group $S_{3}$ is $6$. On the other hand, every element $x \in S_{3}$ satisfies $(1\ 2)x^2 (1\ 2)x^2 = e$, and $6 \nmid 4$.

