Exponent of Schur multiplier need not divide exponent of finite group

Statement
It is possible to have a finite group $$G$$ such that the exponent of the Schur multiplier $$M(G) = H_2(G;\mathbb{Z})$$ does not divide the exponent of $$G$$ itself.

Related facts

 * Schur multiplier of finite group is finite and exponent of Schur multiplier divides group order
 * Product of exponent of finite group and exponent of its Schur multiplier divides order of group