It is known that the set of all transpositions generate the symmetric group.
In the book "Finite Fields" by Lidl and Neiderreiter, in the context of permutation polynomials, they discuss in one of the proofs that it is sufficient to consider the transpositions of the form $(0a)$, where $a\in \mathbb{F}_q^*$ and $0$ is the zero of the field $\mathbb{F}$ for generating the symmetric group $S_q$, because $(bc)=(0b)(0c)(0b)$.
Now, $0$ is usually not considered a part of the set of symbols on which the symmetric group is defined. What do they mean by the 'transposition'$(0a)$ here and how is the multiplication operation defined? I think the multiplication of two 'transpositions' here corresponds to composition of two polynomials associated to the 'transpositions' in the finite field. Any hints Thanks beforehand.
The fact is the normal transposition defined for symmetric groups, except that the symmetric group is defined for the elements of the finite field. The multiplication of transpositions is the normal composition of permutations. The confusion arose because of the faulty thinking that the set of all transpoitions is a minimal generating set of the symmetric group, which is not so.