Theorem SG2
If F and F' are G-congruent 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-1εS(F',G).
Similarly, we can check that, if kεS(F',G), then h = g-1okogεS(F,G).
Since h g-1ohog, k = gohog-1,
so each kεS(F',G) can be written in the
form gohog-1 for some hεS(F,G).
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