# Unit

• Aug 23rd 2012, 11:36 AM
peter1991
Unit
Hello Forum! I have the following problem, which is related to the ring of quaternions, I tried but I had no luck in making it, I hope I can help:

Let I be the ring of integers Hamilton quaternions and define:

$N : I\rightarrow{}Z\ with\ N(a+bi+cj+dk)=a^2+b^2+c^2+d^2$

(The N is called norm)

Prove that an element of R is a unit if and only if it has norm +1. Addition show that $I^x$ is isomorphic to the quaternion group of order 8.

Thanks
• Aug 23rd 2012, 09:08 PM
ModusPonens
Re: Unit
I don't understand the question. the quaternions form a division ring, which means every element except zero has inverse, and thus is a unit.
• Aug 24th 2012, 12:44 AM
Vlasev
Re: Unit
ModusPonens, I think it's because $I$ is the quaternions with integer coordinates. It is similar to the Gaussian integers.

Suppose that $x$ is a unit of $I$. That is, there exists a $y\in I$ such that $xy = yx = 1_I$, where $1_I = 1$ is the identity of $I$. Next, $N$ is a map from $I$ into the non-negative integers. You should prove that $N(ab) = N(a)N(b)$ where the multiplication on the LHS is in $I$ and the multiplication on the RHS is in $\mathbb{Z}$. Then you have

$N(xy) = N(x)N(y) = N(1) = 1$

Can you take it from here?
• Aug 24th 2012, 12:47 AM
Deveno
Re: Unit
he means, presumably, the ring of quaternions with integer coefficients.

what you need to do is show the norm N is multiplicative. that is, for integral quaternions q,q', N(qq') = N(q)N(q').

then if q is a unit in R, we have qp = 1, for some integral quaternion p. so N(q)N(p) = N(qp) = N(1) = 1.

now N(q) and N(p) are integers, so they are units of Z, so N(q) = 1 or -1, since those are the only units of Z.

but N(q) ≥ 0, so N(q) = 1.

this means that precisely one of a,b,c, or d is ±1, that is the units of R are:

{1,-1.i,-i,j,-j,k,-k}.