Let G be a group and let H be a subgroup of G which has exactly two distinct cosets. Let C = {H ′ ⊂ G | H ′ =?
Let G be a group and let H be a subgroup of G which has exactly two
distinct cosets. Let
C = {H ′ ⊂ G | H ′ = gHg^ −1 for some g ∈ G}.
How many elements does the set C have?
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C must have exactly one element. That is, if H has exactly two distinct cosets, then H is a normal subgroup (i.e. invariant under conjugation).
Proof:
If g is an element in H, then trivially, gHg^-1 = H since any subgroup is closed under multiplication by its own elements.
Let g be an element not in H, so that H ≠ gH. It follows that the only two cosets of H are H and gH. Furthermore, if g is not in H, then H ≠ Hg. However, since there are only two cosets and Hg ≠ H, then it must be true that
gH = Hg
It is therefore true that
gHg^-1 = H
Thus, for any choice of g,
gHg^-1 = H
which means that C has exactly one element. QED