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
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.