Subgroup Questions containing 8 elements If $Q$ be the subgroup of $H^*$ generated by the elements $i$, $j$, $k$. Let $N$ = {1, -1}. I am confused on what $H^*$ is exactly. Is it just an arbitrary subgroup of the units of Q, because that is what our other subgroups denoted with the '*' where. I need to show that $Q$ contains 8 elements......
 A: I'm not familiar with every type of notation in this regard. For example, sometimes $\mathbb R^* = \mathbb R \setminus \{0\}$. Sometimes $^*$ indicates a dual space. It depends on the context.
In your case, I assume you are talking about matters related to quaternions.
If you take the linear span of $\{1, i, j, k\}$ over $\mathbb R$, you get a version of $\mathbb R^4$ which has a sensible multiplication defined on it, but it's skew and I guess $\mathbb H^*$ is a multiplicative group without the zero.
But, just looking at the elements $\{i, j, k\}$ with $i^2 = j^2 = k^2 = ijk = −1$ you can generate an 8 element group by only including their unscaled products. It is the subgroup of $H^*$, of finite order cyclic elements.
Addendum 1: On the $^*$ notation thing, yes, if $A$ is a group, according to Serge Lang, in his book, Algebra 3rd Ed, $A^*$ is often used to denote the group of units of $A$. In the case of sets like
$$\mathbb K = \mathbb Q,\mathbb R,\ \mathbb C \text{ and } \mathbb H,\quad \mathbb K^* = \mathbb K \setminus \{0\}$$
as zero is the only non unit. H is sometimes used to mean the Hurwitz quaternions
$$H = \{a + bi + cj + dk \,|\, a, b, c, d \in \mathbb Z\ \text{ or } a, b, c, d \in \mathbb Z + 1/2\}$$
which is complicated. Note here that $H \neq \mathbb H$. However, the Lipschitz quaternions are
$$L = \{a + bi + cj + dk \,|\, a, b, c, d \in \mathbb Z\}$$
that is, just the simple integer span. In this case, where $Q$ is the group generated by non scaled products of $\{i, j, k\}$, $Q = L^*$ is a non trivial use of this notation.
Addendum 2: The units of H:
$$H^* = Q \cup \{\tfrac{1}{2}(\pm 1, \pm i, \pm j, \pm k)\}$$
To get an inverse pair, it must be a conjugate pair. Conjugate by the quaternion definition. Here, $s, t, u, v$ are the signs, which can be either positive or negative freely in any combination.
$$(\frac{s1}{2}, \frac{ti}{2}, \frac{uj}{2}, \frac{vk}{2}) \text{ inverts } (\frac{s1}{2}, \frac{-ti}{2}, \frac{-uj}{2}, \frac{-vk}{2})$$
The usual mathematical way of writing this
$$(\pm\frac{1}{2}, \pm\frac{i}{2}, \pm\frac{j}{2}, \pm\frac{k}{2}) \text{ inverts } (\pm\frac{1}{2}, \mp\frac{i}{2}, \mp\frac{j}{2}, \mp\frac{k}{2})$$
can be confusing.
