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 is an abelian group, and , then there is an element such that , and .

We will prove this by induction. Let us assume that in every abelian group of order , if . Let be a (by default normal) subgroup of . If , by the inductin hypothesis, .

Let us now assume but . This implies . As , by the induction hypothesis, . This implies ( shall be simply be referred to as from now on). is hence that element in such that when raised to the power , gives .

Now all we have to prove is . Given below is my original spin on the proof.

We know . And as is prime, can’t have any common factors with it. Hence . This proves there exist integers such that , where . Also, note that if , then , for any .

Let us now assume . Then . Also note that . The two statements imply . This contradicts the assumption that . Now you would ask where was the assumption made?! The answer lies in the fact we said is the desired element which is not equal to , such that when raised to gives . Had been a part of , then .

There’s an extraordinarily powerful trick I’d like to point out and explain here. When you have statements about and , where , then we can make a statement about by virtue of the fact such that .

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

The statement is :if is prime, and , then there is a subgroup of order in .

We will again prove by induction. If , where is a normal subgroup of and , then the statement is true. Hence, let . This again makes . 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 of order smaller than , if and , then there exists an element such that . Second hint: if , then that does not imply . Moreover, it is not necessary that .*

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