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 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 establish 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.