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