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
.