A new proof of Cauchy’s theorem

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 $\gcd(o(N),p^\alpha)=1$ .