 Solution

On the complex vector space $\mathrm{CF}(G)$ of class functions, define the linear operatorThis is a complex-linear map and satisfies $T^2=I$. It is well-defined on class functions, because inversion of conjugacy classes takes an entire conjugacy class $K$ to another conjugacy class $K^{-1}$.

Use first the basis consisting of the indicator functions $1_K$ of the conjugacy classes. The linear operator $T$ permutes this basis by $1_K\mapsto 1_{K^{-1}}$. A permutation matrix has trace equal to the number of fixed basis vectors, soThe right-hand side counts the real conjugacy classes.

By character orthogonality and completeness, the irreducible characters are another basis of $\mathrm{CF}(G)$. Part (a) gives $T\chi =\overline{\chi }$. The pointwise complex conjugate of an irreducible character is again an irreducible character: conjugate the representing matrices, and conjugate any hypothetical proper invariant subspace back to contradict irreducibility. Thus $T$ also permutes this basis, andHere $\chi =\overline{\chi }$ means exactly that $\chi$ is a real-valued character. Equality of the trace in these two bases provesThis is the real character and real conjugacy class counting theorem. The linear operator used in the proof is $f\mapsto f\circ \mathrm{inv}$; on the irreducible character basis it agrees with pointwise complex conjugation, although pointwise complex conjugation itself is not a complex-linear map on all class functions.