Let $G$ be a finite group & defined a one-one map $T$ in $G$ such that $$aT=a^{-1}\cdot aJ$$ where $J$ is an automorphism of $G$ with a property $aJ=a$ only if $a=e$.
I can prove that $T$ is well-defined and one-one. Also given that the group is finite, I can also prove that $T$ is surjective by pigeonhole principle. But, Is there an alternative for the obvious?
(For example, Let $S_x$ be a function of some finite group $U$ i.e $aS_x = xax^{-1}$ for any element $a$ in $U$. I can prove that $S_x$ is a surjection over $G$ by making use of the definition of Surjection i.e by proving that every element of $U$ has a pre-image. And With little calculations we can find that for every element $a$ in $U$ there exists a pre-image $x^{-1}ax$ in $U$ under the map $S_x$.)
Referring the example, can I prove the surjection of $T$ over $G$ without using the pigeonhole principle i.e Given an element of $G$ can I define the pre-image like in the example to prove the surjection of $T$ over $G$?
Thank you
I think the answer to your question is no.
Consider for example the case where $G$ is infinite cyclic, and $J$ is the automorphism $g \mapsto g^{-1}$. Then $J$ is an automorphism such that $gJ = g$ if and only if $g$ is the identity.
In this case, the map $T$ satisfies $aT = a^{-2}$, so $T$ is injective but not surjective.
Hence there is no formula or expression that you can use to deduce surjectivity of $T$ in the finite case; you will need a different argument. As you mention, in the case where $G$ is finite you can use the pigeonhole principle to conclude that an injective map $G \rightarrow G$ is surjective.