For beginners in Group Theory, the basic method to prove that a subgroup
is normal in a group
is to show that “left coset = right coset”, i.e.
for all
. Variations of this method include showing that
,
, and so on.
This basic method is good for proving basic questions, for example a subgroup of index two is always normal. However, for more advanced questions, the basic method unfortunately seldom works.
A more sophisticated advanced approach to showing that a group is normal, is to show that it is a kernel of a homomorphism, and thus normal. Thus one often has to construct a certain homomorphism and show that the kernel is the desired subgroup.
Example: Let
be a subgroup of a finite group
and
, where
is the smallest prime divisor of
. Show that
is normal in
.
The result above is sometimes called “Strong Cayley Theorem”.
Proof: Let
act on
by left translation.
,
.
This is a group action since
, and
.
This action induces a homomorphism
. Let
.
for all
, i.e.
for all
. In particular when
,
. This means that
. So we have
.
Suppose to the contrary
, i.e.
. Let
be a prime divisor of
.
We also have
![[G:\ker\sigma]=[G:H][H:\ker\sigma]=p[H:\ker\sigma]](https://s0.wp.com/latex.php?latex=%5BG%3A%5Cker%5Csigma%5D%3D%5BG%3AH%5D%5BH%3A%5Cker%5Csigma%5D%3Dp%5BH%3A%5Cker%5Csigma%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
By the First Isomorphism Theorem,
. By Lagrange’s Theorem,
, i.e.
. This implies
. Finally,
implies
.
However,
implies
which implies
.
This is a contradiction that
is the smallest prime divisor of
. Thus,
and therefore
is a normal subgroup.
This proof is pretty amazing, and hard to think of without any hints.