This is a sketch of the proof borrowed from the book "Representations and characters of groups" by Liebeck (p264).

The irreducible character of G are linearly indepdent over as functions from G to , which follows that X is non-singular.

Let be the complex conjugate of matrix X. For each irreducible character , it is known that is also an irreducible character. Thus, we can construct a permutation matrix P induced from such that .

Similarly, for each conjugacy class , the entries in the column of X corresponding to are the complex conjugates of the entries in the column of X corresponding to , where denotes the conjugacy class of its representative , and the conjugacy class of . Similarly, we can construct a permutation matrix Q induced from such that . It follows that . Thus, we see that P and Q have the same trace. Recall that trace of a permutation matrix corresponds to the number of fixed points by the permutation. Therefore, the number of irreducible real-valued characters (corresponding to the fixed points of the first permutation) is the trace of P, and the number of self-inverse conjugacy classes (corresponding the the fixed points of the second permutation) is the trace of Q, which are equal.

Note that iff each is conjugate to , i.e., for some . Then, , , etc. Since G is of odd order, we see that (verify this). Thus, only the identity is conjugate to its inverse. Hence, the self-inverse conjugacy class is {1}. It follows that the number of irreducible real-valued characters is only one by (1).(2) Show that in a group G of odd order, no element other than the identity

is conjugate to its inverse. Deduce that in a group of odd order, the trivial character is the only

irreducible character that is real-valued.