Binary Operation on a Group Problem. Let $\langle Q^{*} \times \mathbb{Z}, * \rangle$ be an abelian group where the binary operation $*$ is defined by
$$(i,j)*(h,k)=\left(\frac{ih}{3},j+k-1\right)$$
Find the value of $(a,b)$ in the equation $(a,b)=(-6,3)^{-1}* (-8,4)^{2}$.
I was able to find that the identity element is $(3,1)$ and the inverse of $(u,v) \in Q^{*} \times \mathbb{Z}$ is $ \left( \frac{9}{u},2-v \right)$. I'm stuck because I don't know what $(-8,4)^{2}$ is. Any ideas?
 A: Perhaps it is worth noting how such an exercise is constructed. It is a trick called transport of structure. Identifying the particular instance of the general trick can be useful in solving such exercises.
Consider the set $A = \mathbb{Q}^{*} \times \mathbb{Z}$. This is clearly a group under the natural operation $$(i,j) \circ (h,k) = (i h, j + k).$$
Consider the bijection $f : A \to A$ given by
$$
f(i, j) = (i/3, j - 1).
$$

Now if $(B, \circ)$ is any group, and $f : B \to B$ is a bijection, it is immediate to see that setting
$$
b_{1} * b_{2} = f^{-1}(f(b_{1}) \circ f(b_{2}))
$$
we obtain a group operation on $B$ such that $f : (B, *) \to (B, \circ)$ is a group isomorphism.

In your case this yields precisely
\begin{align*}
(i, j) * (h, k)
&=
f^{-1}((i/3, j - 1) \circ (h/3, k - 1)))
\\&= 
f^{-1}((i h)/9, j + k - 2)
\\&=
((i h)/3, j + k - 1),
\end{align*}
as $f^{-1}(a, b) = (3 a, b+1)$.

Addendum
Just to see how this helps in your case, and denoting by $x^{\ominus 1}$ the inverse in the $\circ$ group, you will have
\begin{align*}
f(a, b) 
&= 
f( (-6,3)^{-1}* (-8,4)^{2})
\\&=
f((-6, 3))^{\ominus 1} \circ f((-8, 4)^{2})
\\&=
(-2, 2)^{\ominus 1} \circ (-8/3, 3) \circ (-8/3, 3)
\\&=
(-1/2, -2) \circ (-64/9, 6)
\\&=
(32/9, 4),
\end{align*}
so that
$$
(a, b) = f^{-1}((32/9, 4)) = (32/3,5).
$$
A: $$(-8,4)^2=(-8,4)*(-8,4)=(\frac{(-8)(-8)}{3},4+4-1)=(\frac{64}{3},7)$$
Then, by using the inverse rule $(u,v)^{-1}=(\frac{9}{u},2-v)$, we have
$\begin{align}
(a,b)
&=(-6,3)^{-1}*(-8,4)^2\\
&=(\frac{9}{-6},2-3)*(\frac{64}{3},7)\\
&=(-\frac{3}{2},-1)*(\frac{64}{3},7)\\
&=(\frac{(-\frac{3}{2})(\frac{64}{3})}{3},-1+7-1)\\
&=(-\frac{32}{3},5)
\end{align}$
