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What are the steps that leads me to know the presentation (specifically defining the generators and the minimum set of relations required to define this group)of the quaternion group of order 8? What do I need to know about this group to know its presentation?

Note that I know the following:

A general element in the quaternion group has the following form:

$$q = a + ib + jc + kd$$
Where $i,j,k$ satisfies the following relations:

$$i^2 = j^2 = k^2 = -1 \text{ and } ij = k, jk= i, ki=j \text{ and } ji = -k, kj= -i, ik= -j$$

I have found the following questions here presentation of quaternion group of order $8$ , Presentation of a group question and How can i create a presentation of a group ? but still this did not solve my problem.

Any hep will be greatly appreciated!

Shaun
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Brain
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  • The posts do answer your question. They say that $Q_8$ is not abelian, but every group $Q=\langle a \rangle$ is cyclic and hence abelian. So the minimum number of generators is two, i.e., $Q_8=\langle a,b \rangle$. – Dietrich Burde Sep 07 '21 at 14:33
  • @DietrichBurde what about the minimum # of relations? I am asking how to build the presentation of this group? what should I take into account to do this? what are the organized and clear steps to do this? – Brain Sep 07 '21 at 14:41
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    The steps are to see that one generator is not enough, so take two generators $a$ and $b$. Since $i^2=-1$ we have $i^4=1$, so we may assume $a^4=e$. Now try to find the other relations, for $b$ and between $a$ and $b$. This is well explained in all the posts you have mentioned. It takes here at least three relations, if you follow the steps in their constructions. The usual one is $$Q_8=\langle a,b \mid a^4=e,a^2=b^2,ba=a^{-1}b\rangle, $$ but this is not unique. – Dietrich Burde Sep 07 '21 at 14:48
  • And as your last post cited says "There is no general strategy that you can follow to obtain a short presentation for a given group. " Of course, for the quaternion algebra we can find a "short one" easily. But not for a general case of a group. – Dietrich Burde Sep 07 '21 at 14:54
  • Your "general element in the quaternion group" looks more like a general element of the quaternion algebra. – Andreas Blass Sep 07 '21 at 17:46
  • And so what?@AndreasBlass – Brain Sep 08 '21 at 15:26

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