$i,j,k$ are the things we use for quaternion expression.
To show the set is a subgroup, I'm trying to use a subgroup test, which is $ab^{-1} \in \{\pm1,\pm i, \pm j, \pm k, \frac{\pm1 \pm i \pm j \pm k}{2} \} := A$ for any $a,b \in A$.
If $a$ or $b$ is one of $\pm1,\pm i, \pm j, \pm k$, then any product must be in $A$.
So, I would like to show that if $a$ and $b$ is one of $\frac{\pm1 \pm i \pm j \pm k}{2}$, $ab^{-1} = a \frac{\overline{b}}{||b||^2}=a \overline{b} \in A$.
Since I could not find a mathematical way to show this, I am testing with any pairs of elements in $\{\frac{\pm1 \pm i \pm j \pm k}{2} \}$.
However, I don't think trying all pairs of $a,b \in \{\frac{\pm1 \pm i \pm j \pm k}{2} \}$ is the only way to do this.
Is there a better way?
Hint Since the candidate subgroup $A$ is finite, we can just check that the product of two such elements is again in $A$. (After all, $g^{-1} = g^{o(g) - 1}$.)
So, for $p, q, r, s, p', q', r', s' \in \{-1, 1\}$, consider the general product
$$\left(\pm \frac{p + q i + r j + sk}{2} \right)\left(\pm' \frac{p' + q' i + r' j + s' k}{2} \right) .$$
Picking out the coefficient of the $i$-component of the product, which is determined up to a sign, we get (up to sign) $\frac{1}{4}(p q' + q p' + r s' - s r')$. Since all of the coefficients $p, \ldots, s'$ are $\pm 1$, the coefficient of the $i$-component is a half-integer, and the same argument shows that the coefficients of the $1$-, $j$-, and $k$-components are also half-integers.
Incidentally, this construction realizes the group as the so-called tetrahedral group, which is isomorphic to $A_4$.