Given this relation $Q_8=\left<i,j\mid ij=j^{-1}i, ji=i^{-1}j\right>$. 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?

  • $\begingroup$ You can always just take the Cayley table of $Q_8$ and verify that it satisfies the relations. $\endgroup$ – 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 $i^4=ij^2i$

  • $\begingroup$ 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? $\endgroup$ – 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 $i^4=1$.


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