Question

If K is normal in G, prove that $\mathit{K}\mathbf{=}$kernel$\mathit{\phi }$.

Short answer

It is proved that, $K=$Kernel$\phi$ .

Step-by-step solution

Step by Step Solution Step 1: Referring to parts a and b of the question

• $f_a\left(kb\right)=Kba$
• $ker\phi \subset K$
  

Proving that K=Kernelφ

Suppose Ka is any arbitrary element $ka\in T$

Then,$f_{k^{-1}}\left(Ka\right)=Ka{K}^{-1}$.

Since K is a normal subgroup of G, $Ka{K}^{-1}=aK{K}^{-1}\Rightarrow aK=Ka$.

Therefore,$Ka\in T$

$$\begin{gathered} \phi \left(K\right)\left(Ka\right)=f_{k^{-1}}\left(Ka\right)=Ka{K}^{-1} \\ =Ka \end{gathered}$$

Therefore, $K\in \ker \phi$or $K\subset \ker \phi$.

Since we have already proved in part (b) that role="math" localid="1652769622273" $\ker \phi \subset K$, so from both the results, we can conclude that $K=\ker \phi$.

Hence, it is proved that $K=$Kernel$\phi$ .