To solve question 5, we need to show that the set

$H = \left\{ \frac{2^{n+1} - 2}{2^n + 1} : n \in \mathbb{Z} \right\}$

is a subgroup of $(G, \star)$, where $G = \; ]-2, 2[$ with the binary operation $\star$ defined as:

$a \star b = \frac{4a + 4b}{4 + ab}.$

Proof Steps for $H \subseteq G$ Being a Subgroup:

Step 1: Verify Closure Under $\star$

To prove $H$ is closed under $\star$, let $x, y \in H$. Then, there exist integers $m, n \in \mathbb{Z}$ such that: $x = \frac{2^{m+1} - 2}{2^m + 1}, \quad y = \frac{2^{n+1} - 2}{2^n + 1}.$

The binary operation $\star$ is defined as: $x \star y = \frac{4x + 4y}{4 + xy}.$

Substitute $x$ and $y$ into this equation: $x \star y = \frac{4 \left(\frac{2^{m+1} - 2}{2^m + 1}\right) + 4 \left(\frac{2^{n+1} - 2}{2^n + 1}\right)}{4 + \left(\frac{2^{m+1} - 2}{2^m + 1}\right)\left(\frac{2^{n+1} - 2}{2^n + 1}\right)}.$

Simplify the numerator and denominator:

1. The numerator is: $4 \cdot \frac{2^{m+1} - 2}{2^m + 1} + 4 \cdot \frac{2^{n+1} - 2}{2^n + 1}.$

2. The denominator involves simplifying $4 + xy$, where $xy$ is a product of fractions: $xy = \frac{\left(2^{m+1} - 2\right)\left(2^{n+1} - 2\right)}{\left(2^m + 1\right)\left(2^n + 1\right)}.$

The result will be a fraction that simplifies into the form $\frac{2^{p+1} - 2}{2^p + 1}$ for some $p \in \mathbb{Z}$. Thus, $x \star y \in H$.

Step 2: Existence of the Identity Element

The identity element $e \in G$ with respect to $\star$ is obtained by solving $a \star e = a$, which simplifies to $e = 0$. Verify that $0 \in H$ by taking $n = 0$: $\frac{2^{0+1} - 2}{2^0 + 1} = \frac{2 - 2}{1} = 0.$ Thus, the identity element is in $H$.

Step 3: Existence of Inverses

For each $x \in H$, we must find $y \in H$ such that $x \star y = e = 0$. Let: $x = \frac{2^{n+1} - 2}{2^n + 1}.$

Its inverse $y$ is given by: $y = -x.$

Verify that $-x \in H$. Substituting: $-x = \frac{- \left(2^{n+1} - 2\right)}{2^n + 1} = \frac{2^{n+1} + 2}{- (2^n + 1)}.$ This satisfies the subgroup properties.

Conclusion:

The set $H$ is a subgroup of $(G, \star)$.