First we show t_g(a)^-1gt_a G_x for all g G and then certainly for all g B . Indeed,

t_g(a)^-1gt_a(x) = t_g(a)^-1g(a) = x.

Suppose g G_x. Then g G and so it can be written as a product of elements of B and their inverses. Thus,

g = g_r ··· g₁

with g_i B B^-1.

We will show with induction on r, that g can be written as a product of elements from {t_b(a)^-1bt_a | a T,   b B}.

If r = 1, then g B, and with t_g(x)^-1 = t_x^-1 = e, we have g = t_g(x)^-1gt_x, as required.

Assume, therefore, r > 1. Let j be the maximal index such that

x, g₁(x), g₂ g₁(x), ..., g_jg_{j - 1} ··· g₁(x)

is a path in T with the labels

g₁, g₂, ..., g_j.

Observe that j < r as T has no cycles. Put

a = g_jg_{j - 1} ··· g₁(x).

Then

t_a = g_jg_{j - 1} ··· g₁.

Now consider the element g · (t_h(a)^-1ht_a)^-1, where h = g_{j + 1}. Rewrite this element as

g·(t_h(a)^-1ht_a)^-1 = g_r · · · g_{j + 2} ·t_h(a).

Now repeat the above argument on this element; it also belongs to G_x. As t_h(a) corresponds to a path in the Schreier tree T, we find an expression of the form

g·(t_h(a)^-1ht_a)^-1 (t_h'(a')^-1h't_a')^-1 = g_r · · · g_{j' + 2} ·t_h'(a')

with j' > j.

Thus, we can repeat the argument at most r - j times (the head g_r ··· g_{j + 1} becomes shorter and shorter!). We finish with a Schreier element t_b in the right hand side. However, since the left hand side is a product of g and (inverses of) elements stabilizing x, also t_b belongs to G_x. But this is only possible if t_b = e. Consequently, g is a product of elements from {t_b(a)^-1bt_a | a T,   b B} and their inverses. Hence the theorem follows.