Unprovability of $i^2 =1$ from $\langle i \mid i^4 =1\rangle$ and similar problems

Question

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 $i^2 \neq 1$?

Answer 1

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.

Answer 2

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 $$\textbf{D}_8\ncong \textbf{Q}_8$$