the kernel of a group homomorphism is a normal subgroup
In this entry we show the following simple lemma:
Lemma 1.
Let and be groups (with group operations , and identity elements
and , respectively) and let be a group homomorphism
. Then, the kernel of , i.e.
is a normal subgroup of .
Proof.
Let and be as in the statement of the lemma and let and . Then, by definition and:
where we have used several times the properties of group homomorphisms and the properties of the identity element . Thus, and is also an element of the kernel of . Since and were arbitrary, it follows that is normal in .∎
Conversely:
Lemma 2.
Let be a group and let be a normal subgroup of . Then there exists a group homomorphism , for some group , such that the kernel of is precisely .
Proof.
Simply set equal to the quotient group and define to be the natural projection
from to (i.e. sends to the coset ). Then it is clear that the kernel of is precisely formed by those elements of .∎
Although the first lemma is very simple, it is very useful when one tries to prove that a subgroup is normal.
Example.
Let be a field. Let us prove that the special linear group is normal inside the general linear group
, for all . By the lemmas, it suffices to construct a homomorphism
of with as kernel. The determinant
of matrices is the homomorphism we are looking for. Indeed:
is a group homomorphism from to the multiplicative group and, by definition, the kernel is precisely , i.e. the matrices with determinant . Hence, is normal in .