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

(iii) Since $f\left( e \right) = e’$, $e \in K$, this shows that $K \ne \phi$, now let $a,b \in K$, $x \in G$, $a \in K,b \in K$,
$\begin{gathered} \Rightarrow f\left( a \right) = e’,\,\,\,f\left( b \right) = e’ \\ \Rightarrow f\left( a \right) = e’,\,\,\,f\left( {{b^{ – 1}}} \right) = {\left[ {f\left( b \right)} \right]^{ – 1}} = e’ \\ \Rightarrow f\left( {a{b^{ – 1}}} \right) = f\left( a \right){\left[ {f\left( b \right)} \right]^{ – 1}} = e’ \cdot e’ = e’ \\ \Rightarrow a{b^{ – 1}} \in K \\ \end{gathered}$

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

Now, to show that it is also normal we prove the following:
$\begin{gathered} f\left( {{x^{ – 1}}ax} \right) = f\left( {{x^{ – 1}}} \right)f\left( a \right)f\left( x \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left[ {f\left( x \right)} \right]^{ – 1}}f\left( a \right)f\left( x \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left[ {f\left( x \right)} \right]^{ – 1}}e’f\left( x \right) = {\left[ {f\left( x \right)} \right]^{ – 1}}f\left( x \right) = e’ \\ \end{gathered}$

Therefore, ${x^{ – 1}}ax \in K$, hence the result.