**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.25**

Let $I$ be the ring of integral Hamiltonian Quaternions and define $N : I \rightarrow \mathbb{Z}$ by $$N(a + bi + cj + dk) = a^2 + b^2 + c^2 + d^2.$$(1) Prove that $N(\alpha) = \alpha\overline{\alpha}$ for all $\alpha \in I$, where $\overline{a+bi+cj+dk} = a-bi-cj-dk$.

(2) Prove that $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta \in I$.

(3) Prove that an element $\alpha \in I$ is a unit if and only if it has $N(\alpha) = \pm 1$. Then show that $I^\times \cong Q_8$. [Hint: The inverse of $\alpha$ in the rational quaternions is $\overline{\alpha}/N(\alpha)$.]

Solution: Let $\alpha = a+bi+cj+dk$ and $\beta = x+yi+zj+wk$.

(1)\begin{align*}&\ \alpha \overline{\alpha} = (a+bi+cj+dk)(a-bi-cj-dk) \\=&\ a^2 - abi - acj - adk + bai - b^2i^2 - bcij - bdik + caj \\&\ - cbji - c^2j^2 - cdjk + dak - daki - dckj - d^2k^2 \\=&\ a^2+b^2+c^2+d^2 = N(\alpha),\end{align*} after a bunch of cancellation.

(2) \begin{align*}&\ N(\alpha\beta) = N((a+bi+cj+dk)(x+yi+zj+wk)) \\=&\ N((ax-by-cz-dw) +(ay+bx+cw-dz)i \\&\ +(az-bw + cx+dy)j +(aw+bz-cy+dx)k) \\=&\ (ax-by-cz-dw)^2 + (ay+bx+cw-dz)^2\\&\ +(az-bw+cx+dy)^2 +(aw+bz-cy+dx)^2\\ =&\ (a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + z^2 + w^2)\\ =&\ N(\alpha)N(\beta).\end{align*} We have omitted a lengthy cancellation step.

(3) Suppose first that $\alpha$ is a unit. Then $\alpha\beta = 1$ for some $\beta$, where $\beta$ is an integral Hamiltonian Quaternion. Note from the definition of $N$ as a sum of squares that $N(\alpha) \geq 0$ for all $\alpha \in I$. Now $$1 = N(1) = N(\alpha\beta) = N(\alpha)N(\beta),$$ and $N(\alpha)$ and $N(\beta)$ are both integers. Thus $N(\alpha) = 1$. Now suppose $N(\alpha) = 1$. Then $$\alpha\overline{\alpha} = N(\alpha) = 1,$$ and clearly $\overline{\alpha} \in I$, so that $\alpha$ is a unit in $I$.

Suppose $\alpha \in I$ is a unit. Then $$N(\alpha) = a^2 + b^2 + c^2 + d^2 = 1.$$ If any of the $a,\ldots,d$ is at least $2$ in absolute value, we have a contradiction. Thus each $a,\ldots,d$ is either 0,1, or -1. If more than one has absolute value 1, we have another contradiction; thus at most one of $a,\ldots,d$ is 1. Clearly if none are zero, then $\alpha = 0$ is not a unit, and if exactly one is 1, then $N(\alpha) = 1$, so that $\alpha$ is a unit. Thus $|I^\times| = 8$. Since $I^\times$ is nonabelian and has six elements of order 4, $I^\times \cong Q_8$.