Presentation of the Quaternion Group (6.3.7)

Dummit and Foote Abstract Algebra, section 6.3, exercise 7:

Prove the following is a valid presentation: $$Q_8 = \langle~a,b~\mid~a^2=b^2,~a^{-1}ba=b^{-1}~\rangle$$ Proof: By the first, we have $a^2b^{-2}=1$. However, by the second, we also have $a^2b^{-2}=a^2(a^{-1}ba)(a^{-1}ba)=ab^2a$, so that $ab^2a=1$, then $a^{-2}=b^2=a^2$ so $a^4=1$. Now, $a^2=b^2⇒a^4=1=b^4$.

Prove by induction that $b^ka=ab^{-k}$. The case holds for $k=1$ by the presentation, and $b^{k+1}a=bb^ka=bab^{-k}=ab^{-1}b^{-k}=ab^{-(k+1)}$. By extension, $ab^k=b^{-k}a$. We can see by repeated multiplication the left by $a$, that generally $a^ib^k=b^{\pm k}a^i$.

Attempt to observe any element with a reduced width $n$ greater than 2, necessarily in either the form $...a^{e_{n-2}}b^{e_{n-1}}a^{e_n}=...a^{e_{n-2}+e_n}b^{\pm e_{n-1}}$ or $...b^{e_{n-2}}a^{e_{n-1}}b^{e_n}=...b^{e_{n-2} \pm e_n}a^{e_{n-1}}$, a contradiction in any case. So every element can be reduced to the form $a^xb^y$ or $b^xa^y$ (with $0 ≤ x,y ≤ 3$), the latter of which can be ignored as $b^xa^y=a^yb^{\pm x}$. Using the relations we have been given and have established, we can easily establish equivalency classes among these sixteen candidates to provide a maximum of eight distinct elements (e.g. $ab^3=aa^2b=a^3b$). Since $i,j∈Q_8$ fulfill the relations presented above for $a,b$ and $\langle~i,j~\rangle = Q_8$, we have the presentation validated.$~\square$