Question

Let $f:G\to H$ be a homomorphism of groups and let$K=\left\{a\in G| f\left(a\right)=e_H\right\}$ . Prove that K is a normal subgroup of G .

Short answer

It is proved that K is a normal subgroup of G.

Step-by-step solution

Required Theorem

Theorem 8.11: The following conditions on a subgroup N of a group G are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$ for every $a\in G$, Where $a^{-1}Na\subseteq N$$\left\{a^{-1}Na| n\in N\right\}$
3. $aN{a}^{-1}\subseteq N$for every$a\in G$ , Where $aN{a}^{-1}\subseteq N$$\left\{aN{a}^{-1}| n\in N\right\}$
4. $a^{-1}Na\subseteq N$ for every$a\in G$.
5. $aN{a}^{-1}\subseteq N$for every$a\in G$ .

Proving that  is a normal subgroup of  

It is given that $f:G\to H$ is a homomorphism of groups.

And $K=\left\{a\in G| f\left(a\right)=e_H\right\}$, where $e_H$ is the identity element in H .

Now for any arbitrary $g\in G$, we have to show that $gk{g}^{-1}\subset k$.

For any $a\in k$, and from the definition of homomorphism, we can write:

$f\left(ga{g}^{-1}\right)= f\left(g\right)f\left(a\right)f{\left(g\right)}^{-1}$

Since $f\left(a\right)=e_H$,

$\begin{array}{c}f\left(ga{g}^{-1}\right)= f\left(g\right)e_{h}f{\left(g\right)}^{-1}\\ =f\left(g\right)f{\left(g\right)}^{-1}\\ =e_H\end{array}$

Since $e_H\in K,$$ga{g}^{-1}\in k$, which implies, $ga{g}^{-1}\subseteq k$.

Hence, from the above result and theorem 8.11, it is proved that K is a normal subgroup of G.