Kernel of Homomorphism

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 $$f$$.

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 $$G$$.


(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.