# Proving the order of quaternion group is 8

Given this relation $$Q_8=\left$$. I want to show $$Q_8$$ has order $$8$$.

I first tried to prove $$i^4=1$$, but I got stuck. Can anyone give me constructive hint or suggestion?

• You can always just take the Cayley table of $Q_8$ and verify that it satisfies the relations. – freakish Oct 9 at 14:29

First note that because $$ji = i^{-1}j$$, in any string of $$i,j$$s, we can always 'move' the j to the right by modifying the power of $$i$$. So, we can presume that any element can be represented by $$i^mj^n$$, where $$m,n \in \mathbb{Z}$$.

Note that $$i^2 = i(jij) = (iji)j = j^2$$.

Note that since $$iji=j$$, we have $$i^{-1} j^{-1} i^{-1} = j^{-1}$$ and so $$i j^{-1} i = j^{-1}$$.

Note that $$i^4 = i^2 j^2 = ii jj =i(ij)j = i (j^{-1} i) j = (i j^{-1} i) j = j^{-1} j = e$$, and similarly for $$j$$.

In particular, in the expression $$i^mj^n$$ we can take $$m,n \in \{0,...,3\}$$. Hence the order is finite. Because $$i^m j^n = i^m i^{-2}i^2 j^n = i^m i^{-2}j^2 j^n = i^{m-2}j^{m+2}$$, we can further take $$m \in \{0,1\}$$, hence the order is at most $$8$$.

Note: A comment is in order about the order, so to speak.

To show that the order is exactly $$8$$, you would need to find a group that satisfies the presentation rules and has order $$8$$. See https://math.stackexchange.com/a/866048/27978 for an example of how you could do that.

First, show $$i^2=j^2$$, then look at $$i^4=ij^2i$$

• I managed to show $i^{2}=j^{2}$ and $i^{4}=e$ with a help of your suggestion. But how do I show there are only $8$ elements? – aloevera Oct 9 at 15:22

It may be useful to try to break the group representation down further. Canonically, the term $$j^{-1}i$$ is represented as an element unique from $$i$$ and $$j$$. Let's call this element $$k$$. What do you notice about the products $$ij$$, $$jk$$ and $$ki$$? Moreover, what are $$i^2$$, $$j^2$$ and $$k^2$$, and do each of these constructions look familiar? That may help you in showing why $$i^4=1$$.