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Suppose we are given the following presentation of the quaternion group:

$Q_8 = \langle i, j, k \ | \ i^2 = j^2 = k^2 = ijk\rangle$

Is it obvious that $i^4 = 1$?

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  • $\begingroup$ Are the relations supposed to be $$\color{red}{i^2=j^2=k^2=ijk=-1}?$$ $\endgroup$ – Shaun Feb 15 at 22:48
  • $\begingroup$ Oh, wait: Tietze transformations take care of that. Nevermind :) $\endgroup$ – Shaun Feb 15 at 22:50
  • $\begingroup$ @Shaun In general group, it is not clear what "$-1$" is supposed to be; if you mean for it to be just a new symbol which stands for $i^2$ etc., then you gain nothing new. If you mean for it to be a symbol satisfying $(-1)^2=1$, then that trivializes the problem. $\endgroup$ – Wojowu Feb 16 at 10:35
  • $\begingroup$ Well, "$-1$" would just be another generator (despite it being two symbols, I guess) but, as I said above, Tietze transformations would get rid of it, since, in particular, $-1=i^2$, so the transformation of an elimination of a generator would apply; $-1$ is written as a word in the other generators. I think we're saying the same thing, @Wojowu. Thank you for clarifying my second comment here if so. $\endgroup$ – Shaun Feb 16 at 12:12
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From $k^2=ijk$ we get $k=ij$, and from $i^2=ijk$ we get $i=jk=jij$. Hence $i=jij=jjijj=j^2ij^2=i^2ii^2=i^5$, so $1=i^4$.

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