For any group $G$ any $g \in G$, let $L_g:G \rightarrow G$ be the function $L_g(h)=gh$ and $R_g:G \rightarrow G$ be the function $R_g(h)=hg$. Let $S_G$ be the group of permutations of the group $G$ considered only as a set.
Then prove that $g \rightarrow L_g$ is a group homomorphism. What is the kernel of this homomorphism?
Doubt:
Can someone clarify what the question is trying to ask. The notation seems wrong to me. Now clarified
My attempt:
$L_g(h_1h_2)=g(h_1h_2)$ and $L_g(h_1) \cdot L_g(h_2)=(gh_1) \cdot (gh_2)$
How do I show $L_g(h_1h_2)=L_g(h_1) \cdot L_g(h_2)$
What is the kernel of this group homomorphism?
I think it is $G$ because $e$ is the identity of $G$ and $e$ gets mapped to $L_e=G$