> [!NOTE] Theorem (Normal Form of Quaternion Group Elements) > Every element of the [[Quaternion Group|quaternion group]] can be written uniquely in the form $a^ib^j$ with $0\leq i\leq 3$ and $0\leq j\leq 1$. ###### Proof of Normal Form of Quaternion Group Elements By definition, we have $Q_{8} = \langle a, b \mid a^4 =\text{id}, b^2 = b^2, bab^{-1}=a^{-1}\rangle.$ An element $W\in Q_{8}$ is a word over $a,b$ i.e. a finite product of $a,a^{-1},b,b^{-1}$ in any order. Since $a^4=1$ and $a^2=b^2$, we have $b^4=1$. As $a^{-1}=a^3, b^{-1}=b^3$, we suppose the exponent of $a,b$ in the word is non-negative. Since $bab^{-1}=a^{-1}$ we have $ba=a^{-1}b=a^3b.$ Thus we may swap powers of $a$ and powers of $b$ in $W$ so that all $a$ appear on the left and all $b$ appear on the right. Thus $W=a^ub^v$ where $u$, $v\geq 0.$ Write $v=2r+j$ where $j=0$ or 1. Then $\begin{aligned}&W=a^ub^{2r+j}\\&=a^u(b^2)^rb^j\\&=a^u(a^2)^rb^j\\&=a^{u+2r}b^j.\end{aligned}$ Moreover, $a^4=1$. Write $u=2r+j=4w+k$ with $0\leq k\leq 3$ gives $a^{u+2r}=a^k$. Thus $W=a^kb^j$ with $0\leq k\leq 3$ and $0\leq j\leq 1$. It remains to show that no two of the elements $a^kb^j$ are equal. We do this using the following homomorphism $\rho$. $A=\begin{pmatrix}i&0\\0&-i\end{pmatrix},\quad B=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \in \text{GL}_{2}(\mathbb{C})$ It is easy to check that $A^4=I_2,\quad A^2=B^2,\quad BAB^{-1}=A^{-1}$hence it follows from the [[Universal Property of Group Presentations|fundamental theorem of group presentations]] that there is a unique homomorphism given by $\rho:Q_8\to\mathrm{GL}_2(\mathbb{C}),\quad\rho(a)=A,\quad\rho(b)=B.$ Consider $\text{Im}(\rho)$ which is a subgroup of $\text{GL}_{2}(\mathbb{C})$. Note that $|\text{Im}(\rho)| \leq |Q_{8}|=8.$ Also $A$ has order $4$, and $\langle A \rangle \subset \text{Im}(\rho)$. Thus $4 \mid |\text{Im}(\rho)|$. Moreover $B\in \text{Im}(\rho)$ and $B \not \in\langle A \rangle$. Hence $|\text{Im}(\rho)|$ is a multiple of $4$ and strictly greater than $4$, so it is at least $8$. Thus $\text{Im}(\rho)\geq 8$ and so $\text{Im}(\rho)=8=Q_{8}$.