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This question is related to Can I derive $i^2 \neq 1$ from a presentation $\langle i, j \mid i^4 = j^4 = 1, ij = j^3 i\rangle$ of Quaternion group $Q$?

I know I'm going too far but let me just ask...

1) Is indeed $\langle i \mid i^4 =1\rangle$ a presentation for the cyclic group of order 4?

2) If so, then do we just assume that $i^2 =1$ or should we prove that $i^2 =1$ cannot be derived from $i^4 =1$, i.e., the unprovability of $i^2 =1$ from $i^4=1$?

For the below questions I need answers only if the latter is the case in the above question 2.

3) Then how do we rigorously prove the unprovability?

4) As for $$\langle i, j \mid i^4 = j^4 = 1, ij = j^3 i\rangle = Q $$ (Quaternion group), how do I prove unprovability of $i^2 =1$ from the given presentation? Or how do I even guess that $i^2 \neq 1$?

Community
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le4m
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2 Answers2

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1) yes.

2) can't be derived because there is a model for $i^4=1$ in which $i^2\ne1$, namely, $\{{1,\sqrt{-1},-1,-\sqrt{-1}\}}$.

3) See 2).

4) again, by constructing a model in which $i^2\ne1$. One way is by giving the group table for the quaternions.

Gerry Myerson
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All points in @Gerry's answer are covering the cases completely, but just a point about $4$. Let $i^2=1$ when $$\textbf{Q}_8=\langle i,j\mid i^4=1,j^4=1,ij=j^3i\rangle$$ so we are considering the following presentation: $$\langle i,j\mid i^4=1,j^4=1,ij=j^3i, i^2=1\rangle=\langle i,j\mid i^2=1,j^4=1,ij=j^3i\rangle$$ but if $i^2=1$ so $i=i^{-1}$ and since $j^4=1$ so $j^3=j^{-1}$ and therefore we have $$ij=j^3i\longrightarrow ij=j^{-1}i^{-1}\longrightarrow(ji)^2=1$$ This means that we get: $$\langle i,j\mid i^2=1,j^4=1,(ji)^2=1\rangle\cong\textbf{D}_8$$ But we know $$\textbf{D}_8\ncong \textbf{Q}_8$$

Mikasa
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  • Provided that we in fact do know that the dihedral and quaternion groups are not the same, this is a good argument. But if someone is not familiar with both groups, it won't be convincing. – Gerry Myerson Jun 20 '13 at 10:06
  • @GerryMyerson: Thanks so much. Indeed, mine will be useless if the OP don't know that differences. – Mikasa Jun 20 '13 at 10:09
  • Thanks for your answer. Luckily.. I know D_8 is not Q_8. And the answer uses the given presentation for the dihedral group of order 4. It's just that I was looking for more logic-based proof without using group theory knowledges except only primitive ones. Anyway, your answer will be good in situations where we can facilitate the two conditions that you've used. THANKS! – le4m Jun 20 '13 at 11:59
  • @julypraise: Welcome. I am glad I could help you. – Mikasa Jun 20 '13 at 12:14
  • +1 Babak. I miss you! ;-) – amWhy Jun 21 '13 at 04:06
  • @amWhy: Miss you too :( Busy these days cause of my proposal. – Mikasa Jun 21 '13 at 05:51