 Solution

Assume $T$ is a group homomorphism. For any $a,b\in G$, expanding $(ab{)}^3=a^3{b}^3$ and cancelling the initial $a$ and final $b$ gives the useful identityWe also need a consequence of compatibility with conjugation. Calculate the cube of $gz{g}^{-1}$ in two ways:Cancelling shows $g^2{z}^3=z^3{g}^2$. Thus the square of every element of $G$ commutes with every element of the image of a group homomorphism $J=T(G)$.

Now take any $x,y\in J$. In particular $x^{2}y=y{x}^2$. Applying (1) to $x,y$ and then using this commutation givesThis proves the abelian image of a cubing homomorphism property. Since the image of a group homomorphism is a subgroup, $\boxed{T(G)\text{ is an abelian group}}$.