Let $K$ be a subgroup of a group $G$. Let $T$ denote the set of all distinct right cosets of $K$ in $G$ and $A(T)$ be the permutation group of $T$. Prove the following statements.
(a) For each $a\in G$, the function $f_a:T\rightarrow T$ given by $f_a(Kb)=K(ba)$ is a bijection.
(b) The function $\varphi:G\rightarrow A(T)$ given by $\varphi(a)=f_{a^{-1}}$ is a group homomorphism whose kernel is contained in $K$.
I have been able to work out part (a) and most of part (b). I am stuck on the last part of (b) where I am asked to verify that $Ker(\varphi)\subset K$. To do this we need to know the identity element of $A(T)$ which is clearly the identity permutation. In this context that should be the $f_e$ mapping. But this gives me that $Ker(\varphi)=\{e\}$ which doesn't seem correct. Also there is a part (c) that says to prove that if $K$ is normal then $Ker(\varphi)=K$ so I'm pretty sure that I'm wrong but I can't seem to make anymore progress. Any help is greatly appreciated.
Let $h\in Ker(\varphi)$. Note that if your $f_{h^{-1}}$ is the identity permutation, then in particular $f_{h^{-1}}(K)=Kh^{-1}=K$ and this implies that $h^{-1}\in K$ and also its inverse does since $K$ is a subgroup.
Conversely, suppose $K$ is normal and $h \in K$. Then $f_{g^{-1}}(Kb)=Kbg^{-1}$, $f_{g^{-1}h^{-1}}(Kb)=Kbg^{-1}h^{-1}$ and $f_{h^{-1}g^{-1}}(Kb)=Kbh^{-1}g^{-1}$ for all $g\in G$. But $Kbg^{-1}h^{-1}=bg^{-1}Kh^{-1}=bg^{-1}K=Kbg^{-1}$ and $Kbh^{-1}g^{-1}=Kbg^{-1}$ by normality of $K$, so you have that $\varphi(hg)=\varphi(gh)=\varphi(g)$, which implies that $\varphi(h)$ is the identity. So $K$ is included in the kernel and using the first inclusion we conclude that $K=Ker(\varphi)$.