The three conjugacy-class indicators of the symmetric group S3

Conjugation sends a transposition $(ab)$ to $(\tau (a)\tau (b))$: at $\tau (a)$ the composite $\tau (ab){\tau }^{-1}$ sends $\tau (a)$ to $\tau (b)$, at $\tau (b)$ it sends it to $\tau (a)$, and it fixes every other point. In particular $(23)(12)(23)=(13)$ and $(13)(12)(13)=(23)$. Thus all three transpositions, and no other permutations, constitute one conjugacy class.

Similarly, $\tau (123){\tau }^{-1}$ sends $\tau (1)$ to $\tau (2)$, then to $\tau (3)$, then to $\tau (1)$, so it is a three-cycle. Directly $(12)(123)(12)=(132)$, so the two three-cycles form one class. Every conjugate of $e$ is $e$. A permutation of three points is either the identity, a transposition, or a three-cycle: fixing two points forces the third to be fixed; fixing exactly one exchanges the other two; fixing none forces a three-cycle by following any point’s successive images. Hence the three listed classes exhaust $S_3$.

The group is finite and these are exactly its three classes by steps 1.1–1.2, so F1 gives the asserted indicator basis over $k$. More explicitly, for every class function $f$ one has $f=f(e){\mathbf{1}}_{C_1}+f((12)){\mathbf{1}}_{C_2}+f((123)){\mathbf{1}}_{C_3}$, because evaluation on each of the three classes picks out its constant value. For example the coefficients $(0,1,-1)$ give the function with values $0$ at $e$, $1_k$ on the transpositions, and $-1_k$ on the three-cycles. This calculation remains valid in characteristic two, where the last two values coincide, while the indicator basis itself stays independent by evaluation on the three disjoint nonempty classes.