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§ McKay's Proof of Cauchy's Theorem for Groups [TODO ]

created 2022-02-03 · last edited 2022-05-30
  • In a group, if gh=1gh = 1gh=1 then hg=1hg = 1hg=1. Prove this by writing hg=hg(hh−1)=h(gh)h−1=h⋅1⋅h−1=1hg = hg (h h^{-1}) = h(gh)h^{-1} = h \cdot 1 \cdot h^{-1} = 1hg=hg(hh−1)=h(gh)h−1=h⋅1⋅h−1=1.
  • We can interpret this as follows: in the multiplication table of a group, firstly, each row contains exactly one 111.
  • Also, when g≠hg \neq hg=h (ie, we are off the main diagonal of the multiplication table), each gh=1gh = 1gh=1 has a "cyclic permutation solution" hg=1hg = 1hg=1.
  • If the group as even order, then there are even number of 111s on the main diagonal.
  • Thus, the number of solutions to x2=2x^2 = 2x2=2 for x∈Gx \in Gx∈G is even, since each solution has another paired with it.
  • Let's generalize from pairs to
  • Reference
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