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.