= Solution
{c}
Assume $T$ is a <group homomorphism>. For any $a,b\in G$, expanding $(ab)^3=a^3b^3$ and cancelling the initial $a$ and final $b$ gives the useful identity
$$
\boxed{(ba)^2=a^2b^2.}\qquad\text{(1)}
$$
We also need a consequence of compatibility with <conjugation>. Calculate the cube of $gzg^{-1}$ in two ways:
$$
gz^3g^{-1}=(gzg^{-1})^3
=T(g)T(z)T(g)^{-1}=g^3z^3g^{-3}.
$$
Cancelling shows $g^2z^3=z^3g^2$. 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 $x^2y=yx^2$. Applying (1) to $x,y$ and then using this commutation gives
$$
yxyx=x^2y^2,
\qquad
xyx=y^{-1}x^2y^2=x^2y,
\qquad
yx=xy.
$$
This proves the <abelian image of a cubing homomorphism> property. Since the <image of a group homomorphism> is a <subgroup>, $\boxed{T(G)\text{ is an abelian group}}$.
Back to article page
