# Kernel of Homomorphism

Definition
If $f$ is a homomorphism of a group $G$ into a $G'$, then the set $K$ of all those elements of $G$ which is mapped by $f$ onto the identity $e'$ of $G'$ is called the kernel of the homomorphism $f$.

Theorem:
Let $G$ and $G'$ be any two groups and let $e$ and $e'$ be their respective identities. If $f$ is a homomorphism of $G$ into $G'$, then

(i) $f\left( e \right) = e'$

(ii) $f\left( {{x^{ - 1}}} \right) = {\left[ {f\left( x \right)} \right]^{ - 1}}$ for all $x \in G$

(iii) $K$ is a normal subgroup of $G$.

Proof:

(i) We know that for $x \in G$, $f\left( x \right) \in G'$.
$f\left( x \right) \cdot e' = f\left( x \right) = f\left( {xe} \right) = f\left( x \right) \cdot f\left( e \right)$ and therefore by using left cancellation law, we have $e' = f\left( e \right)$ or $f\left( e \right) = e'$

(ii) Since for any $x \in G$, $x{x^{ - 1}} = e$, we get

Similarly ${x^{ - 1}}x = e$, gives $f\left( {{x^{ - 1}}} \right) \cdot f\left( x \right) = e'$
Hence by definition of ${\left[ {f\left( x \right)} \right]^{ - 1}}$ in $G'$ we obtain the result

(iii) Since $f\left( e \right) = e'$, $e \in K$. This shows that $K \ne \phi$, now let $a,b \in K$, $x \in G$, $a \in K,b \in K$,

This establish that $K$ is a subgroup of $G$.

Now, to show that it is also normal we prove the following.

Therefore, ${x^{ - 1}}ax \in K$ Hence the result.