# Presentation of the group of Quaternions

Reading Rotman's book on group theory he states that the following is a presentation of the group of quaternions: $$Q' = \langle x,y \mid x^{2}=y^{2}, \, xyx=y \rangle$$

My question is: how does one shown that $$Q=\{\pm1,\pm i,\pm j,\pm k\}\simeq Q'=\langle x,y\mid x^{2}=y^{2},xyx=y\rangle$$

I believe I can answer my own question. Feedback is welcome!

Notice that working directly with the definition of presentation i.e., proving that $$Q\simeq F_{\{x,y\}}/\triangleleft\{x^{2}y^{-2},xyxy^{-1}\}\triangleright$$, isn't fruitful. Therefore, one is lead to use the universal property of free groups which says that exists a homomorphism $$\tilde{f}$$ as described below. The key observation is that one make an educated guess about the correspondence between the generators of $$Q'$$, $$\{x,y\}$$, and $$Q$$.

$$\begin{array}{ccc} y & \mapsto & j\\ x & \mapsto & i\\ \{x,y\} & \rightarrow & Q\\ \downarrow & \underset{\tilde{f}}{\nearrow}\\ F_{\{x,y\}} \end{array}$$

Now the following computations $$\begin{array}{c} \tilde{f}(x^{2}y^{-2})=i^{2}j^{-2}=-j^{-2}=-(-1)^{-1}=1\\ \tilde{f}(xyxy^{-1})=ijij^{-1}=kij^{-1}=jj^{-1}=1 \end{array}$$

lead to the conclusion that $$\tilde{f}$$ factors through $$Q'$$, i.e. $$\tilde{f}=f\circ f'$$

$$\begin{array}{ccc} F_{\{x,y\}} & \overset{\tilde{f}}{\rightarrow} & Q\\ \underset{f'}{\downarrow} & \underset{f}{\nearrow}\\ Q' \end{array}$$

Thus one gets that $$f:Q'\rightarrow Q$$ is a surjective homomorphism since $$\{x\mapsto i,y\mapsto j\}\subset\text{Im}(f)$$.

The fact that $$xy=yx^{-1}\iff yx=x^{-1}y$$ implies that every element of $$Q'$$ can be expressed in the form $$x^{m}y^{n}$$ where $$m,n\in\mathbb{Z}$$. The relation $$x^{2}=y^{2}$$ restricts the elements of $$Q'$$ to $$x^{m}y^{n}$$ where $$m\in\{0,1\}$$ and $$n\in\mathbb{Z}$$. Now \begin{align*} y^{2} & =xyxxyx\\ & =xy^{4}x\\ & =xx^{4}x\\ & =x^{6}\\ & =y^{6}\\ & \implies y^{4}=1 \end{align*}

Hence every element of $$Q'$$ may be written in the form $$x^{m\in\{0,1\}}y^{n\in\{0,1,2,3\}}$$ which implies that $$|Q'|\leq8$$ and it follows that $$f$$ is indeed a isomorphism.

Note: Apologies for the poorly represented commutative diagrams.