Thanks! Fixed.

]]>Thanks for the detailed comment! Re: Cauchy’s theorem, the argument I suggest doesn’t need the structure theorem for finite abelian groups; we only need to know that a finite abelian group of order is a module over , and then we can decompose into Sylow -subgroups using the Chinese remainder theorem. But your inductive argument is even more elementary!

That’s a nice proof of Sylow I+ as well. And yes, thanks for pointing out the typo.

]]>In the proof of Cauchy via the class equation you write “Cauchy’s theorem is straightforwardly true for finite abelian groups by using the Chinese remainder theorem to write a finite abelian group as a direct sum of finite abelian -groups. (And then, to spell it out very explicitly, Lagrange’s theorem implies that any nontrivial element of a finite -group has -power order, so some power of that element has order .).” If I read this correctly (correct me if I don’t) then this assumes that you already know that finite abelian groups are product of finite cyclic groups. However that not necessary since we can prove it directly by induction on the group order. Indeed, let be a finite abelian group with and pick random . If then we’re done, otherwise we pass to which contains an element of order by the induction hypothesis, its pre-image in then has order divisible by .

In fact, using this we can prove Sylow I+ using the class equation (and without using the structure theorem for finite abelian groups). The case is done same as above. Now let . If is abelian then it contains element of order and has a subgroup of order by the induction hypothesis and we take the preimage. If not then let . If then has a subgroup of order by the induction hypothesis. If then has a -Sylow subgroup by the induction hypothesis and has a subgroup of order by the induction hypothesis and we take the pre-image again.

Finally, in the subproof 1 of Sylow III there should be twice instead of .

]]>Two typos: In proof 1 of Cauchy’s theorem, you write floor of n^2/p, when you mean floor of n/p^2. Also in subproof 1 of Sylow III, both instances of G/N_G(P) should be N_G(P)/P.

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