Say I have a group $G$ and know that for every element $g \in G$, the map $T_g(x) = gx$ is permutation of $G.$
I've heard this map called translation by g before.
Cayley's theorem basically proves this true for all groups $G$ as defined above.
I'm looking to show that $T_g$ is not an automorphism of $G$ unless $g =1_G$ (i.e. unless $g$ the identity element of $G$).
Is the idea that $T_g(x) = gx$ only 1-1 maps from $G$ to $G$ if $g$ is $1$ (identity element under multiplication)?
- Mapping from $G$ to $G$ being used by the definition of automorphism
It is bijection, indeed, but cannot be homomorphism in general case.
If $T_g$ is homomorphism, then $gxy=g(xy)=T_g(xy)=T_g(x)T_g(y)=(gx)(gy)=gxgy.$
So $g=1_G$, because $1_G=(gx)^{-1}(gxy)y^{-1}=(gx)^{-1}(gxgy)y^{-1}=g$.