A new proof of Cauchy’s theorem

by ayushkhaitan3437

We will discuss a more direct proof of Cauchy’s theorem than the one given in Herstein’s “Topics in Algebra” (pg.61).

Statement: If G is an abelian group, and p|o(G), then there is an element g\in G such that g^{p}=e_G, and g\neq e_G.

We will prove this by induction. Let us assume that in every abelian group H of order |H|<|G|, if p|o(H)\implies \exists h\in H: h^p=e_H. Let N be a (by default normal) subgroup of G. If p|o(N), by the inductin hypothesis, \exists n\in N: n^p=e_N=e_G.

Let us now assume p|o(G) but p\not| o(N). This implies p|\frac{o(G)}{o(N)}\implies p|o\left(\frac{G}{N}\right). As o\left(\frac{G}{N}\right)<o(G), by the induction hypothesis, \exists (Nb)\in \frac{G}{N}: (Nb)^{p}=n_1bn_2b\dots n_pb=n_1n_2\dots n_p b^p=N. This implies b^p\in N\implies b^{p.o(N)}=e (e_G shall be simply be referred to as e from now on). b^{o(N)} is hence that element in G such that when raised to the power p, gives e.

Now all we have to prove is b^{o(N)}\neq e. Given below is my original spin on the proof.

We know p\not| o(N). And as p is prime, o(N) can’t have any common factors with it. Hence \gcd (p,o(N))=1. This proves there exist integers such that a.p+b.o(N)=1, where a,b\in\Bbb{Z}. Also, note that if b^{p}\in N, then (b^{p})^{z}\in N, for any z\in \Bbb{Z}.

Let us now assume b^{o(N)}=e. Then (b^{o(N)})^{b}.(b^{p})^{a}=e.(b^{p})^{a}\in N. Also note that (b^{o(N)})^{b}.(b^{p})^{a}=b^{a.p+b.o(N)}=b. The two statements imply b\in N. This contradicts the assumption that b\notin N. Now you would ask where was the assumption made?! The answer lies in the fact we said b^{o(N)} is the desired element which is not equal to e, such that when raised to p gives e. Had b been a part of N, then b^{o(N)}=e.

There’s an extraordinarily powerful trick I’d like to point out and explain here. When you have statements about b^a and b^c, where \gcd (a,c)=1, then we can make a statement about b by virtue of the fact \exists z_1,z_2\in\Bbb{Z} such that z_1.a+z_2.c=1.

Now we consider the proof of Sylow’s theorem for abelian groups, which runs along similar lines.

The statement is :if p is prime, p^\alpha|o(G) and p^{\alpha+1}\not|o(G), then there is a subgroup of order p^{\alpha} in G.

We will again prove by induction. If p^{i}|o(N), where N is a normal subgroup of G and i\leq \alpha, then the statement is true. Hence, let p^{\alpha}|o\left(\frac{G}{N}\right). This again makes \gcd (p^\alpha,o(N))=1. The rest of the proof is elementary.

Anti-climax: The extension to Sylow’s theorem is incorrect. Please try to determine the flaw yourself

Hint: the induction hypothesis is “for groups H of order smaller than o(G), if p^{alpha}|o(H) and p^{\alpha+1}\not| o(H), then there exists an element h\in H such that h^{p^\alpha}=e. Second hint: if p^{\alpha}\not|o(N), then that does not imply p^{\alpha}|\frac{o(G)}{o(N)}. Moreover, it is not necessary that \gcd(o(N),p^\alpha)=1.