Presentation of the group of Quaternions

Question

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 $$Q=\{\pm1,\pm i,\pm j,\pm k\}\simeq Q'=\langle x,y\mid x^{2}=y^{2},xyx=y\rangle$$

Answer 1

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.