Theorem SG2
If F and F' are Gcongruent subsets of S, then S(F,G) and S(F',G) are
conjugate subgroups of G.
Indeed, if gεG is such that F' = g(F), then
S(F',G) = gS(F,G)g^{1}.
Proof
Suppose that F' = g(F) for some g in G. Then we also have F = g^{1}(F').
For hεS(F,G), we have h(F) = F, so that we have
gohog^{1}(F') = goh(F) = g(F) = F'.
Thus gohog^{1}εS(F',G).
Similarly, we can check that, if kεS(F',G), then h = g^{1}okogεS(F,G).
Since h g^{1}ohog, k = gohog^{1},
so each kεS(F',G) can be written in the
form gohog^{1} for some hεS(F,G).

