# The order of the Quaternion Group

I'm trying to find an easy way to prove the order of the Quaternion Group is a non-abelian group of order 8 without many computations, anyone has an idea?

Note we can define the Quaternion Group as the group generated by the following matrices $$A= \left( \begin{matrix} 0 & 1 \\ -1 & 0\\ \end{matrix} \right)$$ $$B= \left( \begin{matrix} 0 & i \\ i & 0\\ \end{matrix} \right)$$ Thanks

• First you have to tell us what definition of the quaternion group you are working with. After all, it can be defined to be the nonabelian, nondihedral group of order 8. – Gerry Myerson Apr 13 '13 at 5:59
• @GerryMyerson ok thank you for the remarks – user42912 Apr 13 '13 at 6:01
• @GerryMyerson I think that's good now – user42912 Apr 13 '13 at 6:11
• Now that you have defined the group, you can show the matrices satisfy the relations in @Babak's answer, then you just have to show that those three relations imply the group elements are as Babak gives them. – Gerry Myerson Apr 13 '13 at 6:24
• @GerryMyerson thanks this makes the problem much easier. – user42912 Apr 13 '13 at 6:42

Or maybe you can find the following an interesting point, a very similar to Ittay's simple way: $$Q_8=\langle a,b\mid a^4=1, a^2=b^2, ba=a^{-1}b\rangle=\{1,a,b,a^2,ab,a^3,a^2b,a^3b\}$$ Note that $ba=a^{-1}b$ shows it is non abelian.
• I think $(ab)^2=1$ is wrong. You can use $ba=a^{-1}b$. – Gerry Myerson Apr 13 '13 at 6:07
• I didn't understand why this $ba=a^{-1}b$ implies that this group is non abelian. Thanks again. – user42912 Apr 13 '13 at 6:48
• If the group were abelian, you'd also have $ba=ab$ --- that's what abelian means, right? If you have both $ba=ab$ and $ba=a^{-1}b$, then you have $a=a^{-1}$, right? so you also have $a^2=1$, right? but we don't have $a^2=1$, we only have $a^4=1$, so if we have $ba=a^{-1}b$ then we can't have $ba=ab$, and the group can't be abelian. – Gerry Myerson Apr 13 '13 at 23:29
Depending on the definition you have in mind, the following might work for you: $ij=k$ while $ji=-k$ shows that it is not abelian. Counting the number of elements in $\{\pm 1, \pm i, \pm j, \pm k\}$ shows that it is a group of order $8$.