Common Hyperplane Medians for Random Vectors
Theodore Preston Hill · American Mathematical Monthly · 1988
is a cyclic group, we can choose a generator g for K *. Then, for x E D *, x E N * is equivalent to xgx-1 E K and, for y E N *, we will have y E Kx if and only if xgx-1 = ygy-1 We let q = IN*/K *. Now D is a (left) vector space over K and, for a E K, Da is a K-linear operator on D. From the above, we have DP =-DPm=Da. Thus the minimal polynomial for Da is a divisor of xpm-x and necessarily splits in K[X] into distinct first degree factors so that Da is diagonalizable. Applying this to Dg, for Ek = {x: [g, x] = kx}, we have Eo K and D = YEk, where the sum is direct and taken over all k E K with Ek # 0. Now, if x E D *, gx- xg- kx for some k E K is equivalent to requiring that x belong to N *. Moreover, y E Ek is equivalent to y E Kx. Then each Ek is a K-subspace of dimension 1 and Ek * ig the coset K *x in N *. Hence, dimKD = q. From the structure of finite fields it follows readily that K is a Galois extension of Z. We can identify N *7K * with a subgroup of G(K/Z) and, if J is the fixed field for N */K *, a E J implies xax-1 = a for all x E N*. Then Da is zero on each Ek so that Da = 0 and a E Z. Hence N *7K * G(K/Z), implying dimzK IN*/K*l = q. Combining the results above leads to dimzD = (dimKD)(dimzK) = q2 and dimZC(b) = q for all b E D, b 5 Z. If r = jZj then ID*I =- 1 and IC(b)*1 = rq- 1 for b 4 Z. Thus, if s is the number of conjugacy classes containing more than one element, the class equation applied to D * gives r 1 = (r- 1) + s(rq- 1/rq-1), which implies r q(q- 1) + + r q + 1 must be a divisor of r- 1, giving the desired contradiction.